Sacred Geometry and Arithmancy.

[Squirrel_Dude]

So, skimming through other forums that talk about 3.5e (minmaxboards.com. Libertad) led to my discovery of the Sacred Geometry and Arithmancy Feats. It's best if I don't try to explain them, and instead let you read them.

Sacred Geometry (http://www.d20pfsrd.com/feats/general-feats/sacred-geometry)
Arithmancy (http://www.d20pfsrd.com/feats/general-feats/arithmancy)

[OldTrees1]

Sacred Geometry seems overpowered. It costs 1 skill point per level to gain a pretty decent chance* of getting 2 free metamagic effects without even taking those feats.

*chance not calculated

Arithmancy seems underpowered. It looks like, Spellcraft DC 19+spell level => +1 caster level.

[Raven777]

These are both the weirdest and coolest feats I have ever seen.

[Snowbluff]

Oh God, where is Grod?

TT.TT

The first is OP, but a huge pain in the ass.

[Raven777]

Grod is busy stalling everyone at the table, calculating Sacred Geometry for his next spell. Duh.

[Snowbluff]

Grod is busy stalling everyone at the table, calculating Sacred Geometry for his next spell. Duh.

He's totally against this sort of thing.

I take it that it would be better to have more ranks in the skill to make it more powerful. That would mean it would be even more annoying to deal with. :smalltongue:

[avr]

Also discussed here (http://www.giantitp.com/forums/showthread.php?363778-Sacred-Geometries).

Even if you could make SG work in a reasonable time it'd distract you from everything else happening at the gaming table. Which would cause misunderstandings and confusion with my group at least.

[Snowbluff]

I would combine with with Spellhunter and Magical Lineage for a cantrip. Then, I would quicken what I was casting. That way, the worst I get is a lost swift action.

[caimbuel]

For a math geek Sacred Geometry is quick easy and absurdly OP, you know those types that in 30 secs they can say yes or no. For the average persona it is a time waster and discouraged at my table.

[NamelessNPC]

These feats are obviously for cheating optimizers. Bogging down the game that hard only increases the chances of Orcus

[inertia709]

These feats give me a math boner. :smalltongue:

Anyway, Arithmancy, like others have said, seems like a "meh" feat at best. Even without the spellcraft check, it still wouldn't be nearly game breaking to get +1 CL as a swift action every turn. Personally, I would likely take it as a Sorcerer or Wizard if it weren't for the Spell Focus(Divination) prereq, but as it stands it seems a bit weak.

With Sacred Geometry, the power depends a lot on the probability of the ability working, and of course the mathematical ability (and patience!) of the player. The probability looks difficult to calculate (that'll be my project for the weekend haha), but on the surface seems to be fairly high with a lot of ranks in Knowledge(Engineering). Do note that the final spell level is limited to the highest level of spell that the player can cast, so that this won't lead to Incantatrix-level shenanigans. That being said, it has the potential to add a lot of "sticking power" to a primary caster in that they can get a lot of extra spells that are effectively at the highest level that they can cast. Quicken + Sacred Geometry in particular would be incredibly potent once the caster gets 5th-level and higher spells. In summary, I could definitely see this being broken in the right player's hands at mid to high levels.

Edit: I just noticed that Sacred Geometry effectively gives you access to 2 new metamagic feats as well as the other benefits. Yeah, this one's almost as broken as the Arcanist.

[oxybe]

Echoing spell+Extend Spell.

Basically with this one feat via Sacred Geomancy, you double the spell slots of any spell 3 levels under your highest. as a 7th level wizard if you roll well enough/can math good enough you get to cast each of your 1st level spell slots twice. And it only goes up from there, because who doesn't want 10-12 castings of their low-level utility spells at that level, especially if the second casting is at double duration (hooray extend! as the option to add SG to a spell is done upon casting the spell, not during your preperation).

Note you can also apply non SG-gained metamagic feats to SG'd spells, so your level 5/10/15/20 bonus metamagic can apply to these too.

And as a wizard, it's not hard to get a stupid high spellcraft, allowing you to easily cast most (if not all) your spells at an additional +1 caster level.

[Thanatosia]

Arithmancy seems really stupid. You can just calculate the 'digital root' of any spells you know or are likely to use ahead of time. Since spells don't randomly change names or spell level, you then have a set DC for the spellcraft check with every spell you know or would use. If your Spellcraft modifier is high enough to guarantee sucess and you have nothing else to use a swift action on that round, you just invoke the feat (+1cl is probably never worth gambling a chance of spellfailure).

It's just a +1 caster level feat with a really stupid and completely arbitrary spellcraft check attached. The way the feat is writtain it's like they expect you to do the calculation on the spot when you cast it and possibly be suprised by the spellcraft DC..... I just don't see any reason anyone would actually play it that way.

[Averis Vol]

What jackass thought it would be a good idea to make these feats? It's such a horridly needless waste of time. Agreed, the first is ridiculously strong (because wizards totally weren't strong enough, and don't waste a huge amount of time with their spells anyways) and the second is by no means worth the effort. Geeze :smallsigh:

[Yanisa]

If it didn't had the awfully game slowing down math, Sacrad Geometry is beyond overpowered.


When casting a spell, you can perform the steps below to spontaneously apply the effects of either or both of these metamagic feats, as well as any other metamagic feats you have, to the spell without expending a higher-level spell slot.
So not only can you pick the two meta-magic feats with this feat, if you posses any other meta-magic feat you can add them to the cause.

AND

You can take this feat more than once; each time, select two additional metamagic feats, adding their effects to the list of possible effects you can apply to spells with this ability.
Basically this feats boils down to 2 metamagic feat for the price of 1.

So in a couple of levels every level spell you cast has an effective spell level of 9 due all the FREE metamagic applied.


As soon as I can get my hands on a smartphone app that calculates this without taking 5 hours of the game session, this is going to ruin games. :smallamused:

[Nilehus]

If someone ever tried to math out how 10 numbers can math to 101 in the middle of a fight, I would destroy them.

Using an app would streamline it, but then they'd essentially be getting two free metamagic feats that they could apply spontaneously. So yeah, gonna agree. OP, cumbersome, and really annoying.

Wizards don't need the help anyway.

[Aliek]

Do you have to use all your die tough?

[Alex12]

Not gonna lie, Sacred Geometry looks weird and stupid and I want to try it.

[Nilehus]

Do you have to use all your die tough?

Seems like.


Then roll a number of d6s equal to the number of ranks you possess in Knowledge (engineering). Perform some combination of addition, subtraction, multiplication, and division upon the numbers rolled that gives rise to one of the relevant prime constants.

Doesn't say anything about dropping dice or anything, so I'd say that you have to use all of them. With how many you'd get that could easily cancel each other out ( /x1, +2 -2, etc.), I'd be very surprised if you couldn't reach it with 15 dice.

That wouldn't stop the other players from trying to feed you your own character sheet, though. :smalltongue:

[meschlum]

Math!

Math is go(o)d!


You have at least 2 dice (required for the feat), and you get feats at odd levels, so you can typically get this at level 3+, guaranteeing 3 dice at least.

With 2 dice, you have 16/36 (44.4%) chance of getting Level 1, and 2/36 (5.56%) chance of getting Level 2
With 3 dice, you have 198/216 (91.7%) chance of getting Level 1, 111/216 of getting Level 2 (51.4%), 54/216 of getting Level 3 (25%) and 15/216 (6.94%) of getting Level 4.

So from this non representative sample, it looks like Level 1 is going to be easy to get. Level 2 should be feasible with 4-5 dice, and higher levels will follow.


As another guesstimate, each die is worth 3.5 on average (via sums), so you need 10 dice for the average sum to be close to the Level 4 requirements. Level 9 would require 30 dice, and is therefore never going to be easy - it will require multiplication to reach even with 17 dice (when 9th level spells are available), so it's not going to happen too often.

With 2 dice, the average sum is 7, and Level 1 happens about half the time - not bad.
With 3 dice, the average sum is 10.5, and Level 2 happens about half the time - again, not bad.

So it looks reasonable to assume that Level 4 will work about half the time when there are 9-11 dice, Level 5 is half the time around 12-14 dice, and so forth. This does mean that a Level 7 spell is never going to be a sure thing - if the assumption holds.

[oxybe]

you do know you can multiply the dice, right?

so at level 7, when you roll your 7 ranks worth of d6 and get (1, 4, 5, 5, 4, 5, 1) trying to reach the Prime Constants of an effective spell level of 4: 31, 37, 41, it's not just adding them to get 25, but rather:

"Perform some combination of addition, subtraction, multiplication, and division upon the numbers rolled that gives rise to one of the relevant prime constants."

so with our rolled numbers we actually get

1+(4x5)+(5x4)-5+1
1+20+20-5+1
21+15+1
36+1
37

we hit our effective level 4 spell and cast at the normal slot.

[Yanisa]

Going by the average of 3.5, four dies would get us: 3*3*4*4 = 144, very much overshooting even effective level 9 spells.

So I guess you need 5-8 dies to hit that sweet spot of effective level 9 spells. Imagine at level 7, being able to apply 6 metamagics to a spell without any drawbacks? For 3 feats?

And then there is Calculating Mind (http://www.d20pfsrd.com/feats/general-feats/calculating-mind), to improve the odds? I dunno but its insane to think there are feats to boosts an already broken feat.

[Thanatosia]

Seems like the key is just multiply 2 dice, or 3 for the real big numbers, to get in the general ballpark, then add/subtract to zero in on a key number. If you can't find a way to get close with 3 multiplied dice, then use 2 multiplied dice as an add/subtract. Try to hit the key number the quickest most direct way you can, then try to find a way to make any remainder dice cancel out.

[ericgrau]

Sacred geometry is op and so rewards math nerdiness. There is a penalty for failure but success is almost assured. When it is hard... then you only bog down the game more as a nerdy jerk. You can't use it to cast higher level spells than you normally could which is good. It still gives a 2 for 1 deal on metamagic feats and gives spontaneous metamagic to prepared casters. Meanwhile to spontaneous casters it says, "screw you, have a two round casting time if you want to use this feat, making it effectively worthless for you.". If that wasn't enough, it conserves high level spell slots.

Ya, way overpowered for one feat. Needs to be 4 feats at least, if not 6. But since it bogs down the game so much it is better to outright ban it.

Arithmancy is usually +1 caster level for a feat and a swift action, which is fair. At lower levels the check is less reliable making it weaker then. As long as the player adds it up before his turn it shouldn't bog down the game ad much but it is still a pretty dumb idea for a feat.

[stack]

Wow, I would expect that from and obscure 3rd party that didn't really know what they where doing, but wow. Kinda neat flavor, which I expect matches the theme of the product it was printed in, but terrible clunky mechanics. Bad Paizo, bad! Where is my rolled up newspaper...hard to smack them on the nose with digital media, breaks the screen.

[Slithery D]

What jackass thought it would be a good idea to make these feats?

They were trying to make something with the right flavor for the occult handbook they published a couple of months ago.

[Hamste]

You can also take the feat multiple times for 2 more metamagic feats each time, why ever buy a single meta magic (other than quicken) when you can two for the price of one? Also did anyone notice that in the example you can get one of the numbers for a 9th level spell if you wanted to? I found that a bit funny that even with just 5 points in engineering (at level 17 for some reason) it is possible to make a 1st level spell have 8 more metamagic levels worth of stuff added onto it.

[Alleran]

I noticed that it was mentioned in the other thread linked above that you might be able to use an app to have a computer do the number-crunching. Does anybody know of such an app?

[Saidoro]

There isn't an app for it. There isn't likely to be an app for it. Figuring solutions for Sacred Geometry is NP-Hard, it's unlikely even a good computer could find a solution for more than a dozen dice or so in a reasonable time-frame. It's a rare feat which you can point to and say "our best current understanding of computer science says this is a bad idea".

[caimbuel]

The only limiting action is doing math, for those that are freaky good like my wizard player this is OP, he did 14 tests in last nights game, took upto 1 min each and made it 13 out of 14 times. I allowed him to roll and math it out before his turn. Biggest effect I saw was casting quickened lvl 1's and 2's. Allowed a ton more versatility to a wizard. I honestly cant say they even remotely needed it. He is going to train back out of it.

[Grod_The_Giant]

Oh God, where is Grod?
You rang?

These feats are hilarious. Arithmancy would be pretty quick if you did the math ahead of time, but Sacred Geometry... I think I have a new favorite feat. It's just absurd on so many levels. It might be cool if it were a once per day thing-- then it becomes a sort of last-ditch power option that the whole table can get in on. But every turn... eesh.

[Hamste]

The only limiting action is doing math, for those that are freaky good like my wizard player this is OP, he did 14 tests in last nights game, took upto 1 min each and made it 13 out of 14 times. I allowed him to roll and math it out before his turn. Biggest effect I saw was casting quickened lvl 1's and 2's. Allowed a ton more versatility to a wizard. I honestly cant say they even remotely needed it. He is going to train back out of it.

Why would they use quicken over other things? They could make a bunch unavoidable level 1 daze spells that causes the opponent to fall over and can hit multiple targets. It can also make the dc to resist spells very hard even with low level spells or make them do a lot of damage if you care about that.

[Killer Angel]

Sacred Geometry is needlessly and stupidly complicated... and totally overpowered.
Sadly, from now on, I'll be condemned to see it in so many TO builds that my brain hurts... :smallsigh:

[Snowbluff]

You rang?

These feats are hilarious. Arithmancy would be pretty quick if you did the math ahead of time, but Sacred Geometry... I think I have a new favorite feat. It's just absurd on so many levels. It might be cool if it were a once per day thing-- then it becomes a sort of last-ditch power option that the whole table can get in on. But every turn... eesh.

I mean, it's exploitable. Yay. With the way skills work, there's probably no reason not to take this feat.

Looking at the fluff, it sounds like you're telling the universe that 1 does, in fact, equal 3, through fallacious misuse of calculus. Thanks, math. :smalltongue:

[CommandTortoise]

I just love how you get more power by doing math outside the game. Makes me think we'll end up sending in cereal box tops to gain a couple of GP ingame.

Think there'll be a melee version of it? For every point of Acrobatics you can attempt one shot with a basketball and deal one extra point of damage for every 3 shots that score. Usable Dex mod times per day, obviously. I think that's around the right balance point.

[Squirrel_Dude]

Someone on the Paizo boards posted this late last night.

Calculating Mind (http://www.d20pfsrd.com/feats/general-feats/calculating-mind)

It's a feat chain. Because of course it is, this is Pathfinder after all.

[Hamste]

Someone on the Paizo boards posted this late last night.

Calculating Mind (http://www.d20pfsrd.com/feats/general-feats/calculating-mind)

It's a feat chain. Because of course it is, this is Pathfinder after all.

Eh, once you have enough points in knowledge (engineering) using d8 instead of d6 won't help much.

[Squirrel_Dude]

Eh, once you have enough points in knowledge (engineering) using d8 instead of d6 won't help much.I'm complaining less about the power increase than Paizo's love for feat chains that have one gem with lots of garbage attached to them.

[Snowbluff]

I'm complaining less about the power increase than Paizo's love for feat chains that have one gem with lots of garbage attached to them.Hush, hush.

Welcome to Pathfinder. :smalltongue:

We need a garbage OP mundane feat now. I feel too dirty in the brain, so I need sword dirt. I just got the idea to have a Druid/Synthesist/BeastmorphVivisectionist/ArcherCleric party, and I'm caster'd out.

[Hamste]

I'm complaining less about the power increase than Paizo's love for feat chains that have one gem with lots of garbage attached to them.

Yeah but in this case the garbage is attached to the end so it really doesn't matter, no one has to take it to get to the feat they want.

[Gemini476]

This feat is pretty great in how it requires the player of the crazy math wizard to act as a crazy math wizard, but it's also pretty crazy in how long time it would take once you get past low levels.

I can maybe see it working for Sorcerers, since they need two turns to cast the spells and can math it out in between, but for wizards? Sheesh. I'd personally adjust the casting time to take as many turns as are needed for the player to figure out the puzzle, but I doubt that would be a popular solution. Especially since wasting three turns of actions for a failed spell would be pretty terrible.

What on earth were the designers thinking of when they made the feat?

[Psyren]

Isn't there still a practical limit on how much you can go nuts with this?

"You can apply any number of metamagic effects to a single spell, provided you are able to cast spells of the modified spell's effective spell level."

So you can't, say, try to quicken all your spells at 2nd level, because you need to be able to cast 4th-level spells before you can use quicken, and you can't attempt to Enlarge + Empower + Maximize all your spells until you can cast 6ths etc. (And even when you can do the former, you can only try to autoquicken cantrips until you can cast 5ths, at which point you can try to autoquicken your 1sts etc.)

[DarkSonic1337]

It may be free metamagic up to your normal casting cap, but it is still FREE METAMAGIC!

Throw on extend to everything, throw dazing on all of your damage over time spells, Persistent Spell all those spells 2 below your highest to make saving against them absurdly difficult, ect. What this feat essentially does is turn your low level spells into effects closer to your higher level spells (though obviously not QUITE as good, which is the reason we don't really like to pay full price for most metamagics anyway).

[Erik Vale]

Some have pointed out apps to do it quickly, but I have a better point.
Use apps to do it, then record the results, write them down on a peice of paper like a cheat sheet. Then you can just reference which rolls work and which ones wont.


And it was used successfully in a game 13/14 times. Let that sink in a moment. Assuming quicken spell was used so as to not boost the time [unless you play sorcerer, in which case it's no different to normal], that's a 7% ASF, that you can turn on and off at whim, in order to boost your spells by ludicrous amounts.

Edit: Is there a metamagic combination that makes cantrips into credible threats though?

Final Edit: Maybe. Dazing Highten Acid Splash. Apply the two -1 metamagic adjustment traits [there could be more] and pump your DCs. You're now throwing around 1d3 acid damage orbs of high DC vs Daze, which prevents the enemy from fighting back, only works if you allow hieghtens level adjustment to be nullified while letting it counting as actually being of higher level to count for Dazing. It allows a wizard to 1 on 1 cherry tap things [I killed you, with a cantrip!].

[Yanisa]

Some have pointed out apps to do it quickly, but I have a better point.
Use apps to do it, then record the results, write them down on a peice of paper like a cheat sheet. Then you can just reference which rolls work and which ones wont.

I tried to convince someone I knew to make a program to calculate this, but he convinced me that this feat isn't to hard to use in practice. Especially when you remember certain key calculations to get close to the number you want and if you have too many dice just sort them in such a way they negate each other.

For example, for level 9th spell the key calculation is 4*5*5=100. Then you just need to get a 3, 5 or 9. from the rest of the dice.

So I start to feel this might be playable at the table, in real life, without taking much of the game time. Especially if you start a low level character, then you can fumble with 2 dice and 3 numbers, slowly working your way up each level, learning the key calculation and getting a feel for sorting dice. The numbers and calculation sound daunting, but it might be a lot less work then people think. I seriously want to try to play this, to see how right I am, but I don't really have a wizard lined up as my next character. (And then there is the issue of overpoweredness once you master math...)

[Kalaska'Agathas]

Arithmancy seems like a lot of work (which should honestly just be done before you get to playing - as Thanatosia pointed out, spells don't randomly change names) for the ability to turn a swift action into a +1 CL.

Sacred Geometry, on the other hand...

That feat is pretty spectacular. Practically getting two metamagic for the price of one feat is pretty awesome on its own (I say practically because it seems you can only use them when you're using Sacred Geometry (though since there's no limit on how many times you can use it per day, the real limit seems to be the "You cannot cast a spell at a level higher than your normal casting cap." thing)). Combine that with free spontaneous metamagic, with the only drawbacks being increased casting time (some of the time) and it being limited to your normal casting cap, however? That's not like icing on the cake - that's icing on the cake plus a whole 'nother cake complete with icing and ice cream to go with it.

I mean, seriously, why would you take one of those more situational metamagics when you could nab the feat you want, plus another one, by taking Sacred Geometry? Not to mention the fact that it looks like you could ignore any prereqs for the metamagics you choose for Sacred Geometry (the feat says "select two metamagic feats you do not yet have," not "select two metamagic feats you do not yet have but meet the prerequisites for." after all). Spectacularly good. Stupidly good, even. Why wouldn't you take this?

Oh, yeah, there's some basic arithmetic involved.

Which shouldn't stop you. Yeah, it can slow things down a bit, but really, not as much as you might think. Per Yanisa:


For example, for level 9th spell the key calculation is 4*5*5=100. Then you just need to get a 3, 5 or 9. from the rest of the dice.

Which basically looks like this:


Geometrician: "I'm going to cast a spell with Sacred Geometry, and an effective spell level of nine! I have seventeen ranks in Knowledge (Engineering) so I roll 17d6!"
Geometrician: rolls dice
Geometrician: "I got a pair of 5s and a 4 here, so I'll multiply those together to get one hundred! Now I need to a get one, a three, or a seven, and get all of my other dice to negate each other, or produce ones, which I can then multiply my previous sum by to get my desired number...okay, I'll take one of my 4s and these two 1s and a 2, which make 0 (since 4-(2+1+1)=0), and these two 6s make another zero (6-6=0), and these two 4s (4-4=0), and these two 3s and this 6 (6-(3+3)=0), and these two 1s (1-1=0), and that leaves me with this 3 which added to my (5*5*4) from before equals one hundred three, one of my prime constants!"

Nota Bene: It took me 18 seconds to do the above calculation, no joke. Just pair off or otherwise combine the dice in front of you, and bam, there you have it.

So yeah, you slow the game down a little bit (or not at all, if you're allowed to pre-roll) and have a small chance of failure, but in the end, get free metamagic, spontaneous metamagic, and two free metamagic feats in one fell swoop. Not quite Leadership, but still quite powerful.

[3WhiteFox3]

Isn't there still a practical limit on how much you can go nuts with this?

"You can apply any number of metamagic effects to a single spell, provided you are able to cast spells of the modified spell's effective spell level."

So you can't, say, try to quicken all your spells at 2nd level, because you need to be able to cast 4th-level spells before you can use quicken, and you can't attempt to Enlarge + Empower + Maximize all your spells until you can cast 6ths etc. (And even when you can do the former, you can only try to autoquicken cantrips until you can cast 5ths, at which point you can try to autoquicken your 1sts etc.)

1. As above it's still free metamagic.
2. You can take SG multiple times with each SG being equivalent to two metamagic feats.
3. You can use metamagic feats without requiring prerequisites.
4. Spontaneous Metamagic

I've run through about a dozen attempts and finding a solution has been trivial each time, taking at most five minutes. My main concern is trying to find out if a roll even has the possibility of success.

On the plus side, this actually opens up several builds that I'd been wanting to try out. On the minus side, it's cheesy and will probably not be open at most tables (if not for balance reasons, then because people will probably use it without a good grasp on doing the math quickly.

[caimbuel]

Want some tips, save odd dice rolls for the end as all final numbers are odd. And odd * even is even, even * even is even, so you need to save atleast 1 odd to make the final number odd. The other feat to make it d8's instead of 6'ss is a waste, plain and simple, if your knowledge in engineering is capped you should have 95%ish chance per the couple hundred rolls we have done. This from someone who does calculus in his head. And my players are math geek/roleplay nerds.

[Squirrel_Dude]

For those who were wondering, Yes, people have already written scripts to do the Sacred Geometry math for you (http://paizo.com/threads/rzs2rb6y&page=4?This-feat-seems-pretty-OP-and-pretty-convoluted#171). Arithmancy, too. (http://paizo.com/threads/rzs2rb6y&page=4?This-feat-seems-pretty-OP-and-pretty-convoluted#173) Both are in Python.

Here (http://repl.it/languages/Python) is a free, in browser Python interpreter.


Oh, and the metamagic feat that no one has brought up yet in relation to this feat. Dazing, Persistent, Empowered, and Extend are all nice, but don't forget Echoing Spell. (http://www.d20pfsrd.com/feats/metamagic-feats/echoing-spell-metamagic)

[nedz]

These feats are just Countdown for 3.5.

I don't like them because they are completely metagamey, also some players would find them easy to use — other players, not so much.

[AnonymousPepper]

So my good friend Al wrote up a little program (http://sd.af/geo/) to have a look at this... and after setting it to do 1000 rolls for every combination of skill ranks and spell levels, it turns out that 12 ranks in Knowledge (Engineering), no Calculating Mind necessary, the chance of coming out with a valid solution is 100% for every effective spell level (http://i.imgur.com/VglJXiQ.png).

All credit to GitP user alcarithemad for this one.

Of course, the primes can be extrapolated further for higher spell levels, in the case of getting higher-level spell slots. Not sure what the required ranks would be for those.

[Necroticplague]

Hmm....When you roll dice, do you have to use all the results? it seems like it gets a fair bit less ridiculously hard if you can just select which dice to use.

[AnonymousPepper]

Hmm....When you roll dice, do you have to use all the results? it seems like it gets a fair bit less ridiculously hard if you can just select which dice to use.

Yes, it's all the dice.

[alcarithemad]

Yes, it's all the dice.

And it doesn't hurt your ability to succeed the rolls at all.

Here's the source for sd.af/geo/: https://github.com/AlcariTheMad/geo

[AnonymousPepper]

And it doesn't hurt your ability to succeed the rolls at all.

Here's the source for sd.af/geo/: https://github.com/AlcariTheMad/geo

Showoff. :smallannoyed:

[Ravingdork]

It seems imperfect. I tried a level 2 spell with 3 ranks and it worked. I tried it with Calculating Mind and it failed.

Seeing as Calculating Mind still lets you use d6s, the latter attempt should have succeeded as well.

[Squirrel_Dude]

Notes on using the program:
1. I wish that the program displayed the dice results even it failed to find the appropriate result.
2. Calculating Mind doesn't replace all the d6s with d8s. It's a player choice to how many d8s or d6s that you can use, but that isn't represented in the program.

[Killer Angel]

If the typical player needs a script to see the effects of a certain feat, then that feat is badly written.

[alcarithemad]

Notes on using the program:
1. I wish that the program displayed the dice results even it failed to find the appropriate result.
2. Calculating Mind doesn't replace all the d6s with d8s. It's a player choice to how many d8s or d6s that you can use, but that isn't represented in the program.

1: Sure, done.
2: Calculating mind is so useless I didn't see the point of allowing mixing.

This takes a pretty simplistic approach to solving it, so for low ranks you may be able to find solutions where it fails.

[Ravingdork]

This takes a pretty simplistic approach to solving it, so for low ranks you may be able to find solutions where it fails.

So it isn't mathematically perfect then? Doesn't that defeat the purpose of automating it in the first place?

[Snowbluff]

A more perfect program would be much more involved, like the difference between a program faking intelligence and an AI, right?

And it doesn't hurt your ability to succeed the rolls at all.

Here's the source for sd.af/geo/: https://github.com/AlcariTheMad/geo

Hm... neat. I wish I knew about programming.

Is there a way I can get this as an app for an iPhone and/or Android device. That seems like the perfect thing to use your extra devices for at a table.

[alcarithemad]

So it isn't mathematically perfect then? Doesn't that defeat the purpose of automating it in the first place?

It isn't, but it's close enough, especially for mid-high level play where you actually have the spell level limits to use it effectively.


It seems imperfect. I tried a level 2 spell with 3 ranks and it worked. I tried it with Calculating Mind and it failed.

Seeing as Calculating Mind still lets you use d6s, the latter attempt should have succeeded as well.

Using d8s when you have less than 5 ranks is worse than using d6s.

[DarkSonic1337]

Wouldn't negating numbers be a simple as

(N-N)(A+B+C+....)+(Prime Constant)?

For example, my random result of 17d6s gave me 4,5,4,1,2,5,6,3,6,1,2,1,4,3,3,3,2. I want to hit 101, 103, or 107.

4x5x5+1+(3-3)x(whatever numbers are left in any combination)=101. This took me longer to roll than to calculate lol.

[Kantolin]

You know, in practice this is actually a lot faster for me than I expected it to be - although granted, I'm pretty good at doing math calculations.

Just look for your target numbers, get hazily near it, and then get rid of your extra dice.

Perhaps the most effective part of it is that 0 * [All the rest of your numbers] = 0. So if your goal is say 11, and you roll 15 dice, get 11 (5 + 6?) then subtract things until you get a 0 and multiply all the rest of the stuff by zero.

So if you get... I dunno 4 2 3 5 1 6 3, then go (6+5) + ((3-3) * 4 * 2 * 1), resulting in (6+5) + 0.

With that, I actually usually can get results in under a minute if not immediately. On my attempts of using 6-7 dice aiming for a +3 total boost, I actually literally never failed and it generally didn't take me anywhere near as long as I thought it would. Although conceivably, if I rolled all ones or something, I could end up with it not working. Usually I take a number that's close to an easy multiple (19 is close to 20, 31 is close to 30, etc) and then finaggle myself to the number (20 - 1 or 20 + 3 or 20 + 9 all work).

Of course, I tend to be pretty quick with math and quick with my turns in general. Several of my friends take forever to use regular attack rolls when buffs are up (Usually I've become the official buff-tracker for the party, although I can't do this when I DM and I'm most often DM) - so this feat would be frustrating because they couldn't use it at all, while I could use it handily. Let alone people who are poor at math.

(Also, free metamagic whee, so I can't see accepting this feat at all, but I play at a lower power level than many people seem to)

(This also means that the relatively dumb sorceror/cleric that can't actually do math at all can do this, while the genius wizard with maximum ranks in Profession(Mathematician) who is so good at numbers he can in fact calculate pi in his head can't do it... if the player can't.)

Edit: And after trying it for +4 and +5 boosts, good lord this gets tons and tons easier when you have more dice to such an extreme amount that I can't see failing. (In which there is math suggesting just this earlier in the topic, mind you, I'm just surprised to see this in practice :smallsmile: )

[Kalaska'Agathas]

So, I'd love to have someone check me on this, but I wanted to figure out what was the lowest number of dice which would guarantee that you can hit at least one of each set of Prime Constants, if they all come up ones. After fiddling with my dice for a bit, I'm pretty sure that you need at least fifteen dice (and therefore, at least fifteen ranks in Knowledge (Engineering)) to do it. I'm not sure that every possible iteration of fifteen dice will let you hit at least one of each set of Prime Constants, but I'm pretty sure that you could.

On something of a tangent, has which Knowledge (or other skill check) is used for mathematics been established?

[Yanisa]

So, I'd love to have someone check me on this, but I wanted to figure out what was the lowest number of dice which would guarantee that you can hit at least one of each set of Prime Constants, if they all come up ones. After fiddling with my dice for a bit, I'm pretty sure that you need at least fifteen dice (and therefore, at least fifteen ranks in Knowledge (Engineering)) to do it. I'm not sure that every possible iteration of fifteen dice will let you hit at least one of each set of Prime Constants, but I'm pretty sure that you could.

So my good friend Al wrote up a little program (http://sd.af/geo/) to have a look at this... and after setting it to do 1000 rolls for every combination of skill ranks and spell levels, it turns out that 12 ranks in Knowledge (Engineering), no Calculating Mind necessary, the chance of coming out with a valid solution is 100% for every effective spell level (http://i.imgur.com/VglJXiQ.png).

All credit to GitP user alcarithemad for this one.

Of course, the primes can be extrapolated further for higher spell levels, in the case of getting higher-level spell slots. Not sure what the required ranks would be for those.

So, 12 dies should do. Although you could probably get away with 10, maybe 9.

Edit: Effective Spell Levels higher then 9 eh?

Well prime numbers tend to rapidly increase as numbers go on... But in praticse you probably won't notice that unless you go to insane high spell numbers. Sticking with 1 rank per effective spell level + 3 would roughly hold up until like something as 18th spell level, and that point you are beyond epic.

There is a short tidbit of prime numbers per spell level

1st 3 5 7
2nd 11 13 17
3rd 19 23 29
4th 31 37 41
5th 43 47 53
6th 59 61 67
7th 71 73 79
8th 83 89 97
9th 101 103 107
10th 109 113 127
11th 131 137 139
12th 149 151 157
13th 163 167 173
14th 179 181 191
15th 193 197 199
16th 211 223 227
17th 229 233 239
18th 241 251 257
19th 263 269 271
20th 277 281 283


Pardon me for the wonky perspective, I not feeling making a table atm. Here is a list up to 333th spell levels (https://docs.google.com/spreadsheets/d/1qdAmW8DEiSSTZ_L0kb-ljADFRoY0JIyPWEkqE30ielw/edit?usp=sharing)... don't ask why... In spreadsheet format.

[Khosan]

So yeah, you slow the game down a little bit (or not at all, if you're allowed to pre-roll) and have a small chance of failure, but in the end, get free metamagic, spontaneous metamagic, and two free metamagic feats in one fell swoop. Not quite Leadership, but still quite powerful.

Easier than that even. You only need to cancel things out once. You could just multiply all your extraneous numbers by your canceled set, getting you 0. So something like:

5 x 4 x 4 + 3 + (6 - 6) x (Big string of extraneous numbers)

EDIT: D'oh. Someone beat me to it. That's what I get for not reading on to page 3 before replying.

[Snowbluff]

So, should we add this for Echoing on Channel the gift?

[AttilaTheGeek]

Sacred Geometry is the worst breach of Grod's Law I've ever seen. It slows down the game, especially during PbPs, and it gives free metamagics. It does so by forcing the player to do some boring arithmetic, but it doesn't care at all (besides the 13 INT prerequesite) how good the character is at boring arithmetic. I don't know what they were thinking when they wrote it.


Call it Grod's Law: You cannot and should not balance bad mechanics by making them annoying to use. When you do:

The disruptive munchkin ignores it, argues it, or forces the rest of the group to suffer through it. His power remains the same, and he gets more annoying to play with.
The inappropriate powergamer figures out how to circumvent the restriction. His power remains the same.
The reasonable player either figures out how to circumvent the restriction (rendering it moot), avoids the class (turning it into a ban) or suffers through it. His power remains the same and/or his enjoyment goes down.
The new player avoids the class or suffers through it. His enjoyment goes down.

[Snowbluff]

Yeah, I pointed that out way back in page one, but thanks for agreeing. WTH? :smalltongue:

[Firest Kathon]

And it doesn't hurt your ability to succeed the rolls at all.

Here's the source for sd.af/geo/: https://github.com/AlcariTheMad/geo

How well does your program scale for many dice? By my thinking, you need to make a lot of operations to just stupidly try all the options. Though looking at your code, it seems to me you just try the dice in the order they are rolled.

Exhausting the full solution space would require
(number of dice)! * 4 ^ (number of dice - 1) operations, if my thinking is correct.
The first part is the possible permutations of the dice order, the second part the possible permutations of the math operations. For 10 dice, this gives me 10! x 4^9 = 951268147200 operations. On a 3GHz processor, assuming 3 clock cycles per operation, that would be roughly 15 minutes, making it a very unfeasible approach.

Edited to correct mathematical expressions

[AttilaTheGeek]

Yeah, I pointed that out way back in page one, but thanks for agreeing. WTH? :smalltongue:

I saw the OP and had to respond immediately. oops :smallredface:

[Snowbluff]

I saw the OP and had to respond immediately. oops :smallredface:

That's exactly what I did. No worries. :smallsmile:

[Sian]

Those saying that 12 dices are 100% chance of being able to get to 101 / 103 / 107 ... i'd like to see you do it with the (highly unlikely but still possible) 1,1,1,1,1,1,1,1,1,1,1,1 :smallwink: chances might be very high 99.999x%, but not 100%

[Hecuba]

I don't know what they were thinking when they wrote it.

My guess is that they were targeting those of us who find arithmetic enjoyable. I do agree however that the implementation here is likely to slow down the game too much for it to be a practical way to get my arithmetic jollies. I'm also somewhat upset that the advertised geometry of sacred geometry isn't present.

[grarrrg]

Those saying that 12 dices are 100% chance of being able to get to 101 / 103 / 107 ... i'd like to see you do it with the (highly unlikely but still possible) 1,1,1,1,1,1,1,1,1,1,1,1 :smallwink: chances might be very high 99.999x%, but not 100%

All 1's boils down to (I'll save you the trouble of "1+1+1=3")
3*3*3*2*2-1 = 107
14 dice minimum

When dealing with 1's, remember that 3 is the Magic Number, and can get you places faster than 2 or 5 (4 is just two 2's in disguise).


But what about all 6's?
6*6=36
6/6+6/6=2 > 6/2 = 3
> 36 * 3 =108
108-6/6 =107
Only Nine 6's
Huh.

[Snowbluff]

My guess is that they were targeting those of us who find arithmetic enjoyable. I do agree however that the implementation here is likely to slow down the game too much for it to be a practical way to get my arithmetic jollies. I'm also somewhat upset that the advertised geometry of sacred geometry isn't present.

Yeah, I like the idea, but this is a huge violation of Grod's Law.

Also, they should have asked use to make right triangles with the numbers. :smalltongue:

[Yanisa]

Those saying that 12 dices are 100% chance of being able to get to 101 / 103 / 107 ... i'd like to see you do it with the (highly unlikely but still possible) 1,1,1,1,1,1,1,1,1,1,1,1 :smallwink: chances might be very high 99.999x%, but not 100%

Well, the percent chance calculations where done on a thousand rolls. The chance of getting 12 ones is about 0.00000000005%. So yeah, whatever I guess? :smalltongue:

[Eanow]

So the 'minimum dice number' problem has been rattling around in my head since I saw this thread a few hours ago.

A few points I'd like to contribute:

The number of unique rolls for N dice is actually pretty reasonable-

http://mathforum.org/library/drmath/view/56115.html
http://www.mathsisfun.com/combinatorics/combinations-permutations.html

12 dice results in 6188 possible unique rolls. 17 dice is 26334. So the space of 'unique rolls for a number of dice' doesn't grow TOO fast.

We could attack the problem from two angles- the lower bound is pushed up by finding counter examples (dice rolls for which you cannot create one of the levels of primes) and the upper bound can be pushed down by considering trivial solutions which always work- for example, we can determine a certain number of '1's is sufficient to generate any of the primes, and we can remove unwanted dice by multiplying them by zero, and a zero will always be available if you have at least 7 dice. So given that G is enough '1's, we can create that many '1's by a few means- any doubled numbers can be turned into a 1, any single difference can become a 1, etc. As a wild overestimation, I think the upper limit for the minimum number of dice is therefore something like (G*2 + 5)+2 (two dice for each 1 we require, and only up to 5 die can exist for which there is not a free die with the same value, and then any two additional die for the zero to cancel out the unwanted die)

Brute forcing this problem will likely be unreasonable, but a couple of these tricks should reduce the problem space into something that a computer could exhaustively search in a reasonable number of cycles.

So far, we know that the 100% mark is higher than 12.

Edit: Looks like we can get all the prime levels from 14 '1's. So given 35 die, we can always create 14 1's and dispose of the rest via the zeroing method, so the upper limit on the minimum number is 35.

[Raimun]

*Cue Conan O'Brien's Nerd Impression:* I shall defeat them with the power of mathematics!

Really, these Feats should not exist. Apart from clunky mechanics, they favor certain kinds of players for all the wrong reasons.

I mean, what next? Boosting weapon damage every time if you (the player, not the character) manage to jump a set amount meters near the gaming table? Bonus to all further Bardic Music bonuses if you manage to knit a mitten during a game session? Bonus Druidic spells if you prove to the DM that you can operate a sailboat by taking him or her to a sailing trip?

The above examples make as much sense as boosting cast spells if you can do math.

[Coidzor]

Yeah, I like the idea, but this is a huge violation of Grod's Law.

Also, they should have asked use to make right triangles with the numbers. :smalltongue:

Yeah, come to think of it, there really should be more Pythagorean elements to it, shouldn't there...

[Eanow]

Yeah, come to think of it, there really should be more Pythagorean elements to it, shouldn't there...

Agreed.

So I wrote a python script to generate all of the unique combinations of die rolls for N dice:



#!/usr/bin/python

def choose_iter(elements, length):
for i in xrange(len(elements)):
if length == 1:
yield (elements[i],)
else:
for next in choose_iter(elements[i+1:len(elements)], length-1):
yield (elements[i],) + next
def choose(l, k):
return list(choose_iter(l, k))
#uses the transformation of thinking of the list of die rolls as a set of bins
#1 2 3 4 5 6
#**||***|***|*|** 2 1's, no 2's, 3 3's, 3 4's, 1 5, 2 6's
#0123456789012345
#have N+5 positions, and 5 'dividers' in positions 1 through N+5
#Use zero index because of code
dice=14 #number of dice
pos=list(xrange(dice+5)) #list of positions
big_list=choose(pos, 5) #create all unique arrangements
print(len(big_list)) #double check that we got the right number
for divider in big_list:
#turn things back into list of die rolls
#number of 1's
number_1=divider[0]
number_2=divider[1]-divider[0]-1
number_3=divider[2]-divider[1]-1
number_4=divider[3]-divider[2]-1
number_5=divider[4]-divider[3]-1
number_6=(dice+5)-divider[4]-1
print str([1]*number_1+[2]*number_2+[3]*number_3+[4]*number_4+[5]*number_5+[6]*number_6)

Now to just write a generalized prime finder that takes in that list...

[meschlum]

Yay!

More math!

I'll grant that I didn't account for the combinatorial explosion (and multiplication options) which come with more dice, so I was obviously underestimating the ease with which totals can accumulate.

Still, some useful bits for people like me.


If you roll N dice, and can get the total you want with X of them, then you can cancel the rest so long as N - X >= 4. In other words, if there are four or more dice left after you get the total you need, they'll cancel out. This is useful, because it means that if (for instance) 14 dice are enough for the primes you're after, you don't need to check if you can also get those primes with 18 or more dice.


So...

With 6 dice, you can cancel five, meaning that you're left with one die. If that one die is a 3 or a 5, you get a level 1 effect. Therefore, with 5 dice, you will always get a level 1 result if any of the rolls is a 3 or a 5. Having applied this restriction, we can work on checking other options. By cancelling four dice (the limit), we're left with two, which we want to combine to get 3, 5, or 7.

The possible dice pairs are as follows, and we can check every pair.

(1, 1): may be an issue
(1, 2): sum is 3, works
(1, 4): sum is 5, works
(1, 6): sum is 7, works
(2, 2): may be an issue
(2, 4): may be an issue
(2, 6): ratio is 3, works
(4, 4): may be an issue
(4, 6): may be an issue
(6, 6): may be an issue

So if any of the 6 dice is not 1, and there is a 1, then (1, 2), (1, 4), or (1, 6) happens. If all dice are 1, we have (1+1+1)*1*1*1 = 3, so it works.

Therefore, with 6 dice, if there is a 1, 3, or 5, then a Level 1 spell can be generated.

If there is a 2 and a 6, then a Level 1 spell can be generated. Dice can be the same, or twos and fours, or fours and sixes.

If there are three identical dice (equal to N), then N + N/N gives 3, 5, or 7. If any of the remaining dice are identical, they cancel out the rest.

With 6 dice and only two values allowed (2 and 4, or 4 and 6), there is always a triplet. With three dice besides the triplet and two dice values, there is always a pair, which cancels the remainder.


Therefore, every possible combination of 6 dice guarantees that a level 1 spell can be generated.

It could be possible with fewer (5 does limit all rolls to 1, 2, 4, 6 so it could be an option), but 6 is certain and relatively quick to prove. The reasoning used can be extended to pools of 7 or more dice, so this demonstrates that Level 1 spells can always be generated with 6 or more ranks.


Combination checking may be more effective with higher spell levels, as things aren't as simple.

EDIT 2: and we've seen that it takes fourteen 1s to get a level 9 spell. You can get a 1 from a double, via division. So you only need 14 doubles to guarantee a level 9 effect.

With 7 dice, you're sure to get a double
With 8 dice, you could have a triplet
With 9 dice, you're sure to have two doubles (either the triplet is now four of the same, or two doubles)
With 10 dice, you could have a triplet (or quint)
...

So to be certain of getting 14 doubles, you need 33 dice. This uses up 28 of your dice, so five are left. Since five is more than four, the remainder can be cancelled out.

Therefore, with 33 dice, you are guaranteed to get a level 9 spell.

It's very likely you can make it with less (by using the numbers you roll rather than converting them into ones), but that sets an absolute upper bound.

In fact:

A pair of 6s means you just need to multiply by 3, and subtract 1, for four extra pairs: 13 dice plus the pair of sixes, so 15 dice.
A pair of 5s means you multiply by 4 and add 1 (101), for 5 extra pairs. 17 dice.
If you roll a 6 and a 5, you need 5 extra pairs (6 * (5 + 1) * (1 + 1 + 1) - 1), which is 17 dice.

So with 17 dice (which you can get by the time you're casting 9th level spells), you just need a pair of 6s, a pair of 5s, or a 6 and a 5 when rolling to succeed with a 9th level spell.

Which means we're looking at getting a 9th level spell with sixteen dice, all ranging from 1 to 4 (and one extra die which could be anywhere from 1 to 6).


If you roll three 4s, you need three more pairs 4 * (4 + 1) * (4 + 1) + 1 if there are no 5s or 6s - which means 12 dice (9 for the pairs and 3 for the 4s). Since you can cancel 4 or more dice at a time, just cancel the extra 5 dice (including the potential 5 or 6), and you're set.

If you roll three 3s, you need five more pairs 3 * 3 * 3 * (1 + 1 + 1 + 1) - 1 if there are no 5s or 6s - which is 16 dice. If the last die is a 1, 2, 3, or 4, it can replace some of the pairs and cancels out. If it's a 5, you need two pairs (3 * 3 * 3 * (5 - 1) - 1), so you're set. With a 6, you need three pairs (6 - 1 - 1 instead of 5 - 1), so you're still good.

Therefore, you might fail to get a 9th level spell with 17 dice if you have at most one 5 or 6, at most two 4s, and at most two 3s.

This means you have twelve or more 1s and 2s.

With two 2s, you have ten or more 1s and (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) * 2 * 2 - 1 works, so you can cast a level 9 spell. This uses twelve dice, so the extra 5 can be canceled out.
With three 2s, you have 9 or more 1s and (2 + 1) * (1 + 1 + 1) * (1 + 1 + 1) * 2 * 2 - 1 works, using 11 dice. Again, this means it works for 17.
With four 2s, you have 8 or more 1s, and it works again
With five 2s, you have 7 or more 1s, and (2 + 1) * (2 + 1) * (2 + 1) * 2 * 2 - 1 uses 9 dice
With six 2s, you have 6 or more 1s, and the solution for 5 works (it uses four 1s), cancelling out the surplus dice.
...
With nine 2s, you have 3 or more 1s, so the solution for five 2s isn't possible. Replace one of the missing 1s with 2/2 and you use seven 2s, which works (10 dice, so the rest cancels)
With ten 2s, you have 2 or more 1s, so you replace another missing 1 with 2/2: (2 + 2/2) * (2 + 2/2) * (2 + 1) * 2 * 2 - 1, using 11 dice (and nine 2s).
With eleven 2s, you have 1 or more 1s, so you make one more replacement: (2 + 2/2) * (2 + 2/2) * (2 + 2/2) * 2 * 2 - 1, using 12 dice (and 11 2s).
With twelve 2s, you could have zero 1s, in which case you'd have a pair of 3s and a pair of 4s. Either pair can give you a value of 1 (3/3), and the other pair (4-4) cancels everything else. So it works.
With thirteen or more 2s, you have (2 + 2/2) * (2 + 2/2) * (2 + 2/2) * 2 * 2 - 2/2, so it works with thirteen dice. The remaining four dice always cancel out.

So you have at most one 2 and eleven or more 1s if there is a chance that you can't cast a 9th level spell.

With eleven 1s, you also have one 2, two 3, two 4, and a 5 or 6. 3 * 3 * 4 * (4 - 1) - 1 works for a 9th level spell, using 6 dice (and two 1s) and the rest cancels out.
With twelve 1s, the solution above works unless a 3 or a 4 is removed. This removed die can be replaced with 1s, so you use 9 or fewer dice (and six 1s), so the remainder can be cancelled out.
With thirteen 1s, another 3 or 4 is removed 3 * 3 * (1 + 1 + 1 + 1) * (1 + 1 + 1) - 1 is a 10 die solution if both 4s are removed, and similar ones are possible if the 3s are removed or if a 3 and a 4 are removed. In any case the extra dice can be cancelled, so it works.
With fourteen 1s, there is still a 3 or 4 left (one 2 at most, one 5 or 6, and three dice that are not 1). So there is a 12 dice solution (3 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1 + 1) - 1), and the five remaining dice cancel out.
With fifteen 1s, there is a 12 dice solution if there is a 3 or 4 left. If not, then there is a 2 and a (5 or 6), and we can use (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) * 2 * (1 + 1) - 1 as a 13 dice solution. The remaining four dice cancel out, so it works.
With sixteen or more 1s, you can use the solution for fourteen 1s and add (1 - 1) * X for the rest.


Therefore, for all possible combinations of 17 dice, it is possible to get a 9th level spell.

Free metamagic for all. QED.

[Promises Kept]

Hmm, a thought, based on the above post. Since so long as you have four or more dice, you can generate a zero, shouldn't it be possible to figure out similar minimum numbers of dice for 1-6? If you know that, you can take several common combinations of dice to get the needed primes, then use that to push the upper bound downwards. For example, 6, 6, 3, 1 is sufficient for 107. If you can guarantee the ability to make a 6 with X dice, a 3 with Y dice, and a 1 with Z dice, you could always generate a 107 with 2X + Y + Z + 4 dice. It won't get you the true value, but it ought to move the boundary closer.

[Sith_Happens]

Why do I suddenly get the feeling that someone from Paizo is reading this thread and laughing evilly?

[Alleran]

Therefore, for all possible combinations of 17 dice, it is possible to get a 9th level spell.
Okay, my eyes glazed over somewhere near/just after where you started bolding stuff. :smalltongue:

So you need 17 ranks minimum, basically?

[meschlum]

Well, I've solved for the practical approach: with 17 dice, you can hit level 9 spells. Take Heighten Spell, and you're set.


Guaranteeing values other than 0 is more challenging (even 0 took some work, which I hve not demonstrated).


You can always get 1 with 7 dice: you have at least one pair, and N/N gives you 1. The remaining five dice are more than four, so they cancel out.


By linear calculation, you'd need 14 dice to get a 2, which is false: with 9 dice, you always have two or more pairs, and N/N + M/M = 2. The extra five dice cancel out, again.


It's this refined approach which gives 33 dice for level 9 spells, but the actual limit is between 14 (minimum using 1s) and 17 (demonstrated possible).


For reference, the cancellation proof:

Roll 4 dice.

If there is a double (D), multiply the sum of other dice by (D - D).

Otherwise.

If there is a 1 AND two of the other dice are adjacent (2-3, 3-4, 4-5, or 5-6), then the difference is zero. Multiply the last number by zero, and done.

If there is a 1 and none of the other dice are adjacent, since you have 4 dice in all, you rolled 1, 2, 4, 6. 6 - 4 - 2 = 0, so you're set.

Note that the feat granting d8s actually makes you more likely to fail here.

So there are no 1s in the four dice you rolled.

If you rolled two pairs of adjacent numbers, the difference is 0 (e.g. 2-3 and 4-5: 2 + 5 - 3 - 4).

If you do not have two pairs of adjacent numbers, you have 2-3-4-6 or 2-4-5-6. In either case, 6 - 4 - 2 = 0.

Note that the feat granting d8s actually makes you more likely to fail here.

So if you rolled four dice, you are guaranteed to be able to combine them to get a zero. This isn't the case if you use d8s, or requires a lot more analysis to confirm.

With three dice, you could roll 1, 2, 5, which does not cancel out (or give 1, so you can't multiply everything else by 1 and change nothing). Therefore, you need exactly four or more dice to be certain that the values can be cancelled.

[meschlum]

Okay, my eyes glazed over somewhere near/just after where you started bolding stuff. :smalltongue:

So you need 17 ranks minimum, basically?

With 17 ranks, you're sure to get Level 9 (and Level 1 spells, demonstration made earlier). I haven't checked for anything else, but Heighten Spell means you don't really care.

It might be possible to be certain to get 9th level spells with as few as 14 ranks (no less), but proving it is difficult.


So the paranoid approach is to take 17 ranks exactly (why use more?) and Heighten Spell. The "likely to work most of the time anyway" approach involves fewer ranks and no Heighten Spell, though the bonus to save DCs is quite glorious.

[Promises Kept]

So... this is effectively one of the most powerful feats in the game. It allows you to spontaneously apply all metamagic you have for free (up to the max level you can cast), gives you access to two metamagic feats without the need to meet their prerequisites, and effectively makes all of your spell slots 9th-level equivalents. And, as the rest of this thread has proven, it's very reliable, and not particularly difficult to use. Is there any reason for a caster *not* to take this feat, if they can manage the skill ranks?

[Kalaska'Agathas]

So... this is effectively one of the most powerful feats in the game. It allows you to spontaneously apply all metamagic you have for free (up to the max level you can cast), gives you access to two metamagic feats without the need to meet their prerequisites, and effectively makes all of your spell slots 9th-level equivalents. And, as the rest of this thread has proven, it's very reliable, and not particularly difficult to use. Is there any reason for a caster *not* to take this feat, if they can manage the skill ranks?

Not mechanical ones. It does have the potential to slow the game down, a little (depending on how fast you are at basic arithmetic), and it may be too powerful for some tables, so those may be valid reasons not to take it.

[Yirrare]

Since everyone seems to miss it, I would just like to point out the hidden benefit of calculating mind.


[...]Then roll a number of d6s equal to the number of ranks you possess in Knowledge (engineering).[...]


[...]You can use any combination of d6s and d8s that you wish, as long as the number of dice does not exceed the number of ranks you possess in Knowledge (engineering).[...]
I have bolded the relevant parts for clarity. In short, calculating mind gives you the option of which die combination and number of die you wanna use.

But then, to be fair, you will probably want to use as many die as you can along with the elimination method first suggested by DarkSonic1337.

Best Regards
Yirrare

[meschlum]

Further number crunching indicates that Level 1 spells can be done with 5 or more ranks. 4 ranks don't always work: (5, 5, 5, 5) fails, hence the minimum is 5 ranks. Edit: 4 ranks actualy works.

To get a Level 2 spell with only 1s, you use 13 = (1 + 1 + 1) * (1 + 1 + 1 + 1) + 1, for 8 dice. So the lower limit is 8 dice.

With 10 or more dice, you have at least two pairs.

If you have a pair of 4s, you have another pair (which gives 1), and you can use 4 * 4 + 1, with four dice, so there are six left, which cancel out.

So you have one 4 at most.

If you have a single 4 and a pair of 2s, you can use 2 * 2 * 4 + 1, and it works.

If you have a single 4 and a single 2, you can use (2 + 1) * 4 + 1, using six dice. The remaining four dice cancel out.

So if you have a single 4, you cannot have any 2s.

If you have a single 4 and one or more 3s, you can use 3 * 4 + 1, with four committed dice (one pair, one 4, one 3), so it works since you have two pairs or more.

So if you have a single 4, you cannot have any 2s or 3s. This leaves nine or more dice with values of 1, 5, or 6.

If you have four or more 1s, you can use (1 + 1 + 1) * 4 + 1, with five or more dice left, so it works. Therefore, you have at most three 1s.

If you have three 1s, you can use (1 + 1 + 1) * 4 + N/N, where N=5 or N=6, since you have six or more dice which are 5s or 6s, so you have at least two pairs. This uses six dice, and the remainder can be canceled.

If you have two 1s, you have seven or more dice that are 5s or 6s, so at least three pairs. (1 + 1 + P/P) * 4 + P/P uses seven dice, leaving one pair. (P-P) = 0, so the remainder cancels out.

If you have one 1, you have at least three pairs of 5s and 6s, so you have a solution using 8 dice. Your remaining dice are either equal and cancel out, or are 5 and 6, and 6-5 = 1, so that cancels out as well.

So if you have a single 4, you can only have 5s or 6s.

If you have four 6s, you can get 6 + 6 + 6/6, which works. So you have three 6s at most.

If you have four 5s, you can get 5 + 5 + 5/5, which works. So you have three 5s at most.

Since you need to have nine or more 5s and 6s, yuo have at least four of either, so you can always cast a level 2 spell with ten or more dice and one or more 4s.


Now we look at what happens if you have no 4s and ten or more dice. In this case, you have at least three pairs.

If you have a pair of 6s or a pair of 5s, you can do 6 + 6 + 1 (or 5 + 5 + 1), using four dice and cancelling the rest.

So you have at most one 5 and one 6. If you have both, 5 + 6 works with the rest cancelling out, so you have at most one die that comes out as a 5 or a 6.

You therefore have nine or more dice that are 1s, 2s, or 3s.

If you have three or more 3s, you can use 3 * (3 + 3) - 1 with five dice (a pair of 2s or a single 1 will contribute the 1)

If you have two 3s, you can use (3 + 3) * 2 + 1 if you have 1s and 2s, or (3 + 3) * 2 + 2/2 if you have only 2s, or (3 + 3) * (1 + 1) + 1 if you have only 1s. So it always works.

If you have one 3, you can use 3 * 2 * 2 + 1 if you have at least two 2s and a 1, or 3 * 2 * 2 + 2/2 if you have only 2s (and a 5 or 6), so you have one 2 at most (and at least eigth 1s). 3 * (1 + 1 + 1 + 1) + 1 uses six dice, so the rest cancels out.

You therefore have nine or more dice that are 1s or 2s, and one that could be 1, 2, 5, or 6.

If you have a 5 or 6, 6 * (1 + 1) + 1 or 5 * (1 + 1) + 1 means you have at most two 1s (so seven or more 2s). 6 * 2 + 2/2 and 5 * 2 + 2/2 mean you have at most two 2s, so there is always a solution if you have a 5 or 6.

Therefore, you have ten or more dice that are 1s or 2s.

With six or more 2s, you have 2 * 2 * 2 *2 + 2/2, and the rest cancels out, so there is always a solution in this case and you have at least five 1s.

With three or more 2s, you have 2 * 2 * (2 + 1) + 1, and the rest cancels out if you have at least two 1s, which is the case.

With two 2s, you have (1 + 1 + 1) * 2 * 2 + 1, using 6 dice and the remaining 1s cancel out.

With one 2, you have a (1 + 1 + 1) * 2 * (1 + 1) + 1, with only 1s otherwise, so it works.

With no 2s, you have only 1s, and there is a solution with eight 1s, the extra 1 cancels out (multiply by 1).


Conclusion: With 10 or more dice, you are always certain of getting a level 2 spell.

It is possible to be certain with as few as 8 dice, but no less.

The absolute minimum dice pool to be certain of a level 1 result is 4 dice (2nd level spells).

So the minimum dice pool for absolute certainty starts out fairly high and rises quickly (to 8 for level 2) before flattening out (at 17 or fewer dice for level 9), meaning a paranoid caster will never bother with low level spells - by the time you're confident that you can cast level 2 spells without failure, you can probably do a lot better.

[grarrrg]

Further number crunching indicates that Level 1 spells can be done with 5 or more ranks. 4 ranks don't always work: (5, 5, 5, 5) fails, hence the minimum is 5 ranks.

(5-5)*5+5=5
Four 5's can get a 5 and pass Level 1

It's late, I ain't checking all of you other stuff....night...

[meschlum]

(5-5)*5+5=5
Four 5's can get a 5 and pass Level 1

It's late, I ain't checking all of you other stuff....night...

Indeed! Which is why it's hard to get absolute lower bounds in many cases. My mistake.


With 4 dice:

If there is a pair and a 3 or a 5, then grarrrg's approach works.

If there are four 1s, then Level 1 is possible

If there are three 1s, then Level 1 is possible (1 + 2, 1 + 4 and 1 +6, or a pair of 1s and a 3 or 5)

If there is a pair of 1s, then the other two dice are 2, 4, or 6

(2, 2): 2 + 2 + 1
(2, 4): 2 + 4 + 1
(2, 6): 2 + 6 - 1
(4, 4): 4 + 4 - 1
(4, 6): 6 - 4 + 1
(6, 6): 6 - 6/(1+1) (requires all 4 dice)

So there is at most one 1.

If there is a 1 and a pair, then the pair cancels out and the remaining options are 2, 4, 6. Since 1 + 2, 1 + 4 and 1 + 6 all work, Level 1 is always possible if there is a 1 and a pair.


If there are four 2s, then 2 + 2 + 2/2 works

If there are three 2s, then we need to check combinations with 2, 4, and 6 (there is a pair of 2s and a 3 or 5 otherwise).

2 + 2 + 2 + 1 works
4 + 2 + 2/2 works
6 - 2 + 2/2 works

If there is a pair of 2s, then the other two dice are 4s, or 6s since Level 1 is always possible with a 1 and a pair of 2s.

(4, 4): 2 + 2 + 4/4
(4, 6): 6 - 4 + 2/2
(6, 6): 2 + 2 + 6/6

So there is a most a single 2.

If there are three or more 3s (or 5s) then there is a pair of 3s (or 5s) and an extra 3 (or 5), so Level 1 is possible.

If there are two 3s, then the other dice are 2s, 4s, or 6s (at most one of each, since a pair and 3 works)

(2, 4): (2 + 4) / 3 + 3
(2, 6): (6 / 2) / 3 * 3
(4, 6): 4 + 6 / 3 - 3

So there is at most a single 3

If there are two 5s, then the following checks apply

(2, 4): (5 - 4) * (5 - 2)
(2, 6): (6 - 5) * (5 - 2)
(4, 6): (4 + 5 + 6) / 5

So there is at most a single 5

If there are four 4s, we use 4 + 4 - 4/4

If there are three 4s, we need to check combinations with 2 and 6

4 + 2 + 4/4
6 - 4 + 4/4

So there are at most two 4s.

If there are two 4s, we need to check combinations with 2 and 6 (since there is a pair, so 1, 3, and 5 all give Level 1 spells), and only one 2 is allowed.

(2, 6): 2 + 6 - 4/4
(6, 6): 4 + 4 - 6/6

So there is at most one 4.

If there are four 6s, then (6 * 6) / (6 + 6) works

If there are three 6s, we need to check with 2 and 4

6 - 6 + 6/2
6 - (6 + 6) / 4

So there are at most two 6s.

If there are two 6s, we need to check combinations with 2 and 4 (since 1, 3, 5 are excluded and there cannot be pairs of 2s or 4s)

(2, 4): 6 + 6 / (2 + 4)

So there is at most one 6.


Therefore, with 4 dice, Level 1 is always possible if there is a pair.

If there are no pairs, the remaining options are:

4 + 3 - 2 * 1
5 + 3 - 2 - 1
6 + 3 - 2 * 1
5 + 4 - 2 * 1
6 + 4 - 2 - 1
6 - 5 + 2 * 1
(5 - 4) * 3 * 1
(6 - 4) * 3 + 1
(6 - 5) * 3 * 1
6 - 5 + 4 * 1
5 + (4 + 2) / 3
6 / (3 * 2) + 4
6 / (3 * 2) * 5
6 / (2 + 4) * 5
(6 - 3) * (5 - 4)

So all 4 dice combinations work for Level 1.

I stand corrected, Level 1 can be done with 4 dice (and there are proofs for 5 and 6+, so it really is 4+ dice).

With three dice, (1, 6, 6) does not work, so the absolute lower limit is 4.
Recap

Level 1: 4+ ranks exactly (level 2+ needed for the feat)
Level 2: 8+ ranks exactly (level 3+ needed to cast)

Level 9: 17+ ranks guaranteed, could be as low as 14 (level 17+ needed to cast)

[GreyBlack]

I am totally the jackass at my table who would use this feat. This feat is WAY OP and should be banned at every table, as it is way too easy to pull off and way too much for way too little.

[Miss Disaster]

Conclusion: With 10 or more dice, you are always certain of getting a level 2 spell.

It may be possible to be certain with as few as 8 dice, but no less.

The absolute minimum dice pool to be certain of a level 1 result is 5 dice (3rd level spells for prepared casters).

So the minimum dice pool for absolute certainty starts out fairly high and rises quickly (to 8-10 for level 2) before flattening out (at 17 or fewer dice for level 9), meaning a paranoid caster will never bother with low level spells - by the time you're confident that you can cast level 2 spells without failure, you can probably do a lot better.Thanks for number-crunching all that data out, Meschlum. It was an interesting read.

So as a recap to your synopsis that I quoted above .... is there a general guideline for each of the Levels of Spells that has a correlating number total for Knowledge: Engineering ranks in regards to achieving *absolute certainty* with Sacred Geometry?

Ex.
1st Level Spell = 5 Ranks
2nd =
3rd =
4th =
5th =
6th =
7th =
8th =
9th = 17 Ranks

That's 11 Knowledge: Engineering skill ranks between the high and low end ... and 7 spell level to correlate them too. Unless that doesn't work with the stat formula and table spread.

[meschlum]

Thanks for number-crunching all that data out, Meschlum. It was an interesting read.

So as a recap to your synopsis that I quoted above .... is there a general guideline for each of the Levels of Spells that has a correlating number total for Knowledge: Engineering ranks in regards to achieving *absolute certainty* with Sacred Geometry?

Ex.
1st Level Spell = 4 Ranks
2nd =
3rd =
4th =
5th =
6th =
7th =
8th =
9th = 17 Ranks

That's 11 Knowledge: Engineering skill ranks between the high and low end ... and 7 spell level to correlate them too. Unless that doesn't work with the stat formula and table spread.

My pleasure! Math is fun (to me).

(Note that Level 1 can be done with 4 ranks, as revealed by number crunching above - low levels are painful).

(Also note that Level 9 may be possible with fewer than 17 ranks (as few as 14), but the proof (if there is one) becomes harder).

Short answer: no. If there were a way to easily procedurally generate prime numbers, a lot of the current paradigms for password security would be overthrown.

Slightly longer answer: for the limited range of prime factors that are used by the feat, it's possible to do the math. The issue is doing it in a sane amount of time.


I can work on filling in the table, though!

Level 2, revised. We know that Level 2 can be done with 10 or more ranks, and at least 8 are needed. Let's see if eight or more is possible.

If you roll two 6s, two 5s, or two 4s, and have a pair, you get Level 2 with 4 dice (5 + 5 + 1, 6 + 6 + 1, 4 * 4 + 1). The remaining four or more dice cancel out.

If you roll two 6s, two 5s, or two 4s and have no other pairs, the remaining six dice are (1, 2, 3, 4, 5, 6). You can use the 1, and have five dice left, which cancel out.

So you have at most one 4, one 5, and one 6.

If you roll a 5 and a 6, you have six dice left which cancel out and 5 + 6 works.

So you have at most one (5 or 6)

If you roll a 4 and a 6, you have six dice left with values in the 1 to 3 range, so there is a double, and you can get 4 * 4 + 1 with four dice. Your remaining four or more dice cancel out, so you can't have a 4 and a 6.

If you roll a 4 and a 5, the remaining dice are in the 1 to 3 range. If you also roll a 3, 4 * 5 - 3 works (and you have 5 dice left, which cancel out). If you also roll a 2, 4 + 5 + 2 works (and you have 5 dice left, whcih cancel out). If you have six 1s, 4 + 5 + 1 + 1 works.

So one die can be 4, 5, or 6 (or 1, 2, 3), and the remaining seven must be 1s, 2s, or 3s.

If you have two 3s and a 2, the extra dice cancel out.

If you have two 3s and two 1s, the extra dice cancel out.

So if you have two 3s, you can have at most one 1 and the other dice are 3s (or one die at 4, 5, 6)

With seven 3s, you have 3 * 3 + 3 + 3/3 with five dice, and a pair of 3s left, so (3-3) cancels the rest.

With six 3s and a 1, you have 3 * 3 + 3 + 1 with four dice, and the rest cancels out.

So you have at most one 3.

If you have a 3 and a 4, you have six dice with 1s and 2s, so you have a pair. 3 * 4 + 1 works.

If you have a 3 and a 5, you have six dice with 1s and 2s, so you have at least two pairs 3 * 5 + 1 + 1 works, leaving you with two dice worth 1 or 2. If they are identical, they cancel out. If they are 1 and 2, then 2-1 = 1, and that cancels as well.

If you have a 3 and a 6, you have six dice with 1s and 2s, so you have a pair. 3 * 6 - 1 works.

So you have at most one 'wild' die which can be worth 3, 4, 5, or 6, and the remaining are 1s and 2s.

If the wild die is a 4, you have 4 * 2 * 2 + 2/2 and 4 * (1 + 1 + 1) + 1, so there are at most three 1s and three 2s and three dice are not used to get the total. These unused dice are 1s or 2s, so there is a pair, and they cancel out. Since there are at least seven 1s and 2s, a Level 2 spell can always be cast if a 4 is rolled.

If the wild die is a 3, you have 3 * 2 * 2 + 2/2, 3 * (1 + 1 + 1 + 1) + 1, and 3 * 2 * 2 + 1, so there are at most four 1s and three 2s. If there are 1s and 2s, there is at most a single 2 (since 4 dice are used in 3 * 2 * 2 + 1, so the rest cancels out). So if there are 1s and 2s, there must be at least six 1s, and the extra 1s cancel out - so there must be either 1s only or 2s only. In either case, there are at least two dice left, which are identical, so they cancel out as we have a pair.

If the wild die is a 5, you have 5 * 2 + 2/2 with four dice, so there are at most two 2s. You also have 5 * (1 + 1) + 1, so there are at most two 1s. Since there must be at least seven 1s and 2s, a Level 2 spell is always possible in this case.

If the wild die is a 6, you have 6 * 2 + 2/2 and 6 * (1 + 1) + 1, so again a Level spell is always possible.

Therefore, you need all eight dice to be 1s and 2s.

Since the dice are 1s and 2s, any two dice will cancel out (either equal or multiply by 2 - 1).

If you have six or more 2s, 2 * 2 * 2 * 2 + 2/2 is a solution, and the remaining two dice cancel out. So you have at most five 2s, and at least three 1s.

If you have four or more 2s, 2 * 2 * 2 * 2 + 1 works, and you have enough 1s to make it possible. So you have at most three 2s, and at least five 1s.

With three 2s, you get 2 * 2 * (2 + 1) + 1, which works (the remaining 1s cancel out).

With two 2s, you get 2 * 2 * (1 + 1 + 1) + 1, which works (6 dice).

With one 2, you get 2 * (1 + 1) * (1 + 1 + 1) + 1, which works (7 dice).

With no 2s, you have eight 1s, which work (1 + 1) * (1 + 1) * (1 + 1 + 1) + 1

So Level 2 is possible with eight dice, and not with seven (try seven 1s)
The new table is:

1st Level Spell = 4 Ranks (exact)
2nd = 8 Ranks (exact)
3rd =
4th =
5th =
6th =
7th =
8th =
9th = 17 Ranks (could be as few as 14)

So there is a rather steep initial curve, then 9 Ranks cover 7 Levels. And it's not impossible that some intermediate spell levels could require more dice than the high spell levels!

I've got proofs that Level 3 works with 11 dice, but it could be as low as 9. That's for later, though.

[grarrrg]

My pleasure! Math is fun (to me).
[STUFF]

Might I recommend Spoiler Tags? The length is getting a bit much.
My poor, poor scroll wheel!

[meschlum]

Might I recommend Spoiler Tags? The length is getting a bit much.
My poor, poor scroll wheel!

But then how can I inflict math on hapless passers-by?

Done anyway.



If we have a 6 and a (3, 4, or 5), there are 7 or more dice left, so there is a pair. 6 * 5 - 1, 6 * 4 - 1 and 6 * 3 + 1 all work for Level 3.

So if there is a 6, there are only 1s, 2s, and 6s.

If there are five or more 6s, 6 + 6 + 6 + 6/6 is a solution, with at least 4 dice left, so it cancels out. Therefore, there are four 6s at most.

If there are three 6s and a pair, 6 + 6 + 6 + 1 is a solution. With four 6s at most, there are at least five dice that are 1s or 2s, so there are at least two pairs. One pair cancels out the extra, so there is a solution.

If there are one or two 6s, there are at least seven dice that are 1s or 2s. Since 6 * (2 + 1) + 1 is a four die solution (and the remaining five dice cancel out), there is at most one 1 if there are any 2s.

With a 6 and a 1, there are at least six 2s if any 2s are present. 6 * (2 + 2/2) + 1 uses five dice, and the remaining four or more cancel out.

With a 6 and seven or more 1s, 6 * (1 + 1 + 1) + 1 uses five dice, and the remaining four or more cancel out.

So if there are one or two 6s, the other seven or more dice are 2s. 6 * (2 + 2) - 2/2 uses five dice, and the remaining four or more cancel out.

Therefore, a Level 3 spell is always possible with nine dice if one is a 6.

If we have a 5 and a 4, there are seven dice left, so there is a pair, 5 * 4 - 1 is a solution, and uses four dice or less, so the remainder cancels out.

If we have a 5 and a (3 or 5), 5 * (3 + 1) - 1 and 5 * (5 - 1) - 1 are solutions, so there is one 1 at most, and one pair at most among the seven dice.

If there is one pair among seven dice, then (1, 2, 3, 4, and 5) are present, and one is a triplet. This means that the pair can create a 1, and the other 1 can be used as well. With five dice in all (2 for the pair, one for the 1), the remaining four dice can be cancelled, so there is a solution if there is a 5 and a 3 or two or more 5s.

If we have a single 5 and the remaining dice are 1s and 2s, 5 * 2 * 2 - 2/2 is a 5 die solution and the rest cancels out, so there are three 2s at most. 5 * (1 + 1) * (1 + 1) - 1 is a solution as well, with six dice. Since the remaining dice are 1s or 2s, there is a pair among them, and they cancel out. Thus, there are four 1s at most. This gives seven dice other than the 5 at most, and we use nine dice in all, so there is always a solution.

Therefore, with nine dice, a Level 3 spell is always possible if there is a 5 or a 6.

If there are two or more 4s, 4 * (4 + 1) - 1 is a solution. Since we have seven dice spread across 1, 2, 3, and 4, there are at least two pairs, so the solution is possible, using six dice in all. The solution uses five dice if there is at least one 1, allowing the rest to cancel, so the six die solution must have no 1s. This leaves seven dice spread across 2, 3, and 4. If the remaining three or more dice after the solution include a pair, the remainder cancels out and there is a solution. Otherwise, there are exactly three remaining dice (9 dice total) and the remaining dice are 2, 3, and 4, with a solution of 19. If we add 4 * (3 - 2), we get a solution of 23, which is also valid.

Therefore, there is at most a single 4 and the eight or more other dice are 1s, 2s, and 3s.

If there is a 4, 3, 2, 1, then 4 * (3 + 2) - 1 is a four die solution and the remainder cancel out. Therefore, if there is a 4, all three other dice values do not appear.

If there is a 4, 3, 2, then 4 * (3 + 2) - 1 is a five die solution (using one pair) and the remainder cancel out. With a 4, 3, 2, there are at least six other dice spread across 2 and 3, so there is a pair and the solution exists.

If there is a 4, 3, 1, then 4 * (3 + 1) - 1 is a five die solution (using one pair). Again, the pair is possible and the remainder cancels out.

If there is a 4, 2, 1, then 4 * (2 + 1 + 1) - 1 is a seven die solution using two pairs. This leaves at least two dice which are 1s and 2s - if there is a pair, the remainder canels out. Otherwise, there are exactly two dice left, which are 1 and 2, and we can multiply the result by (2-1), so the solution exists.

Therefore, if there is a 4, all the other eight or more dice have the same value.

3 * 3 * 3 - 4 is a four die solution and the remainder cancel out
4 * (2 * 2 + 2/2) - 2/2 is a seven die solution, so there are at least two 2s left, which give 2-2 and cancel the remainder
4 * (1 + 1 + 1 + 1 + 1) - 1 is a seven die solution, and all the remaining dice are 1s, so we can multiply and keep the solution.

Therefore, there is a solution with nine or more dice if any of the dice are 4, 5, or 6.

3 * 3 * 2 + 1 is a solution where the remainder cancel out, so if there are at least two 3s, there cannot be 1s and 2s together.

(3 + 2) * (3 + 1) - 1 is a solution where the remainder cancel out, so if there are at least two 3s and some 1s, there is a single 1 at most.

(3 + 2) * (3 + 2) - 2 is a solution where the remainder cancel out, so if there are at least two 3s and some 2s, there is a single 2 at most.

So if there are two 3s, there are at least eight 3s and one (1 or 2).

3 * 3 * (3 - 3/3) + 3/3 uses seven 3s. If the remaining dice are 3s, the solution exists. If the remaining dice are a 2 and a 3, multiply by 3-2 and a solution exists, so if there are two or more 3s, a solution exists unless there is a 1 as well.

3 * 3 * (3 - 1) + 3/3 uses five 3s and a 1, so the remaining three or more dice are all 3s, and cancel out.

Therefore, there is at most a single 3 with nine or more dice, and eight dice that are 1s or 2s.

3 * (2 + 1) * 2 + 1 is a solution where the remainder cancel out, so if there is a 3 and 1s and 2s, there is either a single 1 or a single 2.

With a single 2, 3 * (2 + 1) * (1 + 1) + 1 is a solution, and the remaining dice are all 1s, so they cancel out.

With a single 1, 3 * (2 + 1) * 2 + 2/2 is a solution using six dice, so there are at least three 2s remaining, and 2-2 cancels them all.

So if there is a 3, all remaining dice are 1s or all remaining dice are 2s.

3 * 2 * 2 * 2 - 2/2 is a six die solution, and the remaining three or more 2s can be cancelled with 2-2
3 * (1 + 1) * (1 + 1 + 1) + 1 is a seven die solution, and the remaining two or more 1s cancel out.

So Level 3 spells can be done with nine dice if any come up 3, 4, 5, or 6.

2 * 2 * 2 * 2 + 2 + 2/2 uses seven dice, and the two or mor remaining are 1s and 2s, which cancel out (2-2, 1-1 or multiply by 2-1 if there are exactly two remaining). Therefore, there are at most six 2s and at least three 1s.

(2 + 1) * (2 + 1) * 2 + 1 uses six dice, and the remaining three or more dice are 1s and 2s, and so include a pair and cancel out. This means that there are at most two 2s, and at least seven 1s.

(2 + 1) * (2 + 1) * (1 + 1) + 1 uses seven dice, and the remaining dice must be 1s since the 2s have been used, so they cancel out. Thus, there is at most a single 2 and at least eight 1s

(2 + 1) * (1 + 1 + 1) * (1 + 1) + 1 uses eight dice, and the remaining dice must be 1s since the 2 has been used. Again, the solution works, and there cannot be any 2s

(1 + 1 + 1) * (1 + 1 + 1) * (1 + 1) + 1 uses nine dice, and the remaining dice must be 1s since no others are allowed.

Therefore, Level 3 spells can always be cast with 9 or more dice, and cannot always be cast with eight or less (try eight 1s)



Level 1: 4 or more Ranks (exact)
Level 2: 8 or more Ranks (exact)
Level 3: 9 or more Ranks (exact)
Level 4:
Level 5:
Level 6:
Level 7:
Level 8:
Level 9: 17 or more Ranks (could be as low as 14)

I expect that Level 6 and Level 8 will be problematic, because there are no convenient shortcuts to get close to the associated primes.

[Miss Disaster]

Level 1: 4 or more Ranks (exact)
Level 2: 8 or more Ranks (exact)
Level 3: 9 or more Ranks (exact)
Level 4:
Level 5:
Level 6:
Level 7:
Level 8:
Level 9: 17 or more Ranks (could be as low as 14)

I expect that Level 6 and Level 8 will be problematic, because there are no convenient shortcuts to get close to the associated primes.

More interesting insight, Meschlum. Thanks again.

Your quoted text above is intriguing in your claim that the impromptu table doesn't actually have consistent and predictable progression (where you mention that 6th & 8th level gets complicated). Makes you wonder if the designer(s) of the feat were aware of that arithmetic quirk.

Still, whether this feat makes it to my gaming table or not, I do appreciate that it has forced me a few others in my group to bone-up on our admittedly rusty math and statistics knowledge.

[meschlum]

More interesting insight, Meschlum. Thanks again.

My pleasure!


Your quoted text above is intriguing in your claim that the impromptu table doesn't actually have consistent and predictable progression (where you mention that 6th & 8th level gets complicated). Makes you wonder if the designer(s) of the feat were aware of that arithmetic quirk.


I'm fairly certain they were not, since the series used is just three consecutive primes for each spell level, rather than anything clever. Besides, the pattern of required ranks goes against what I'd expect from sensible design - you want low level boosts to be easy, so people benefit from the feat early on with minor Metamagic, and high level boosts to be difficult, so there is a reason not to cast all your magic missiles maximized and quickened. And the trend of having low incerases for higher level effects goes against that.



If you have two or more 6s, you have nine dice left, so there is a pair. 6 * 6 + 1 works, so there is one 6 at most (four dice used, so the remaining seven or more cancel out).

If you have two or more 5s, you have nine dice left, so there are two pairs 5 * (5 + 1) + 1 works, so there is one 5 at most (six dice used, so the remaining five or more cancel out).

If you have a 5 and a 6, you have nine dice left, so there is a pair. 5 * 6 + 1 works, so one die can come up as a 5 or a 6 (four dice used, so the remaining seven or more cancel out).

If you have three fours, you have eight dice left, so there is a pair. 4 * (4 + 4) - 1 works, so there are two fours at most (five dice used, so the remaining six or more cancel out).

If you have a 4 and a 6, you have nine dice left (two pairs). 6 * (4 + 1) + 1 works, with six dice used and the remaining five or more cancel out.

If you have a 4 and a 5, you have nine dice left, all of which are 1s to 4s. This means three pairs, and (4 + 1) * (5 + 1) + 1 works, using eight dice. The remaining three dice are all in the 1 to 4 range. If there is a double, they cancel out, so the following options exist:

1, 2, 3: add 3 - 2 - 1
1, 2, 4: multiply by (4 - 2 - 1)
1, 3, 4: add 4 - 3 - 1
2, 3, 4: multiply by 2 + 3 - 4

Therefore, if you have any 4s you cannot have a 5 or 6.

If you have two 4s, you have nine dice left, which are all 1s to 3s.

If you have two 4s and a 2, you have eight dice left (one pair). 4 * 4 * 2 - 1 works, so you can't have a 2 if there are two 4s (five dice used, so the remaining six or more cancel out), and the other dice are all 1s or 3s.

If you have two 4s and a 3, you have eight dice left which are 1s or 3s, so three pairs. 4 * (4 + 3) + 3 is a solution, using four dice (and the remainder cancel out), so there is a single 3 at most, and at least seven 1s. (4 + 1 + 1) * (4 + 1) + 1 uses six dice (the remaining five cancel out) and four 1s, so it is a solution if there are seven or more 1s.

If you have two 4s and only 1s, (4 + 1 + 1) * (4 + 1) + 1 works, and there are at least nine 1s, with the remainder canceling out.

So there is at most a single 4, which is not compatible with having a 5 or 6, so one die can be 4, 5, or 6 and the others are all 1s, 2s, and 3s

If there are three or more 3s, there are eight dice left, one of which can be 4, 5, or 6, so there are two pairs among the seven dice coming up 1, 2, and 3. 3 * 3 * (3 + 1) + 1 is a solution, using seven dice. This leaves four dice, which cancel out. Therefore, there are two 3s at most.

If there are exactly two 3s, there are nine or more dice left, one of which can be 4, 5, or 6. This leaves eight or more dice that can be only 1s or 2s, guaranteeing three pairs.

If there are two 3s and two or more 2s, there are seven or more dice left, of which six are 1s or 2s, and so include a pair. 3 * 3 * (2 + 2) + 1 is a solution with six dice, and the rest cancel out. So if there are two 3s, there is at most a single 2, one 4, 5, or 6, and the other seven or more are 1s.

With two 3s and seven 1s, we have 3 * 3 * (1 + 1 + 1 + 1) + 1 using five 1s, which leaves two 1s to cancel out the rest via 1-1. Therefore, there cannot be two 3s.

If there is one 3, there are nine or more dice that must be 1s or 2s (and one that can be anything but a 3). 3 * 2 * 2 * (2 + 2/2) + 2/2 uses eight dice and seven 2s, so there must be at least three 1s. 3 * 2 * 2 * (2 + 1) + 1 is a solution with less than three 1s, so there must be two 2s at most, and at least seven 1s.

With one 3 and two 2s, we get 3 * 2 * 2 * (1 + 1 + 1) + 1, leaving three or more 1s to cancel the rest. So there is one 2 at most (and eight 1s or more).

With one 3 and one 2, we get 3 * 2 * (1 + 1) * (1 + 1 + 1) + 1, using six 1s. Since there are at least eight 1s, the remaining two or more cancel out via 1-1.

With one 3 and nine or more 1s, we get 3 * (1 + 1) * (1 + 1) * (1 + 1 + 1) + 1, using eight 1s. This leaves at least one 1 and a die that must be 1, 4, 5, or 6.

If the wild die is a 1, they cancel out.

If the wild die is a 4, 3 * 4 * (1 + 1 + 1) + 1 leaves five or more 1s to cancel the rest.

If the wild die is a 5, 3 * (5 + 1) * (1 + 1) + 1 leaves five or more 1s to cancel the rest.

If the wild die is a 6, 3 * 6 * (1 + 1) + 1 leaves six or more 1s to cancel the rest.

So there cannot be any 3s, and we have ten dice that are 1s or 2s, as well as a wild die which is 1, 2, 4, 5, or 6.

If the wild die is a 4, then 4 * (1 + 1) * (1 + 1) * (1 + 1) - 1 is an eight die solution and the three remaining dice are 1s or 2s, so there is a pair and they cancel out. Therefore, there are six 1s or less and at least four 2s. If there is a 4 and three 2s, there are seven dice left (1s or 2s), so there is at least one pair, and we have 4 * 2 * 2 * 2 - 1 as a solution, using six dice and three 2s as well as a pair. The remaining dice cancel out, so the wild die cannot be a 4.

If the wild die is a 5, then 5 * (1 + 1) * (1 + 1 + 1) + 1 is a seven die solution and the remainder cancels out. This means there are at most five 1s, and at least five 2s. If there is a 5 and four 2s, there are six dice left (1s and 2s), so there is at least one pair, and we have 5 * 2 * (2 + 2/2) + 1 as a seven die solution. The remainder cancels out, so the wild die cannot be a 5.

If the wild die is a 6, then 6 * (1 + 1 + 1 + 1 + 1) + 1 a seven die solution and the remainder cancels out. This means there are at most five 1s, and at least five 2s. If there is a 6 and four 2s, there are six dice left (1s and 2s), so there is at least one pair, and we have 6 * 2 * (2 + 2/2) + 1 as a seven die solution. The remainder cancels out, so the wild die cannot be a 6.

Thus all eleven dice must be 1s or 2s.

If there are seven or more 2s, 2 * 2 * 2 * 2 * 2 - 2/2 is a solution, leaving four dice that cancel out. Thus, there are at most six 2s, and at least four 1s.

If there are five or more 2s, 2 * 2 * 2 * 2 * 2 - 1 is a six die solution, and the remainder cancels out. This leaves at most four 2s, and at least seven 1s.

If there are two or more 2s, 2 * 2 * (1 + 1) * (1 + 1) * (1 + 1) - 1 is a nine die solution with seven 1s (which is required). The remaining two dice are 1s or 2s, and cancel out (either as a pair or multiply by 2-1)

If there is a single 2, 2 * (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) - 1 is a ten die solution, with only 1s left, so the remainder cancels out. Thus, there are no 2s.

With eleven 1s, (1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) - 1 is a solution.

Therefore, there is always a solution for level 4 with eleven dice, and there may not be with ten dice (all 1s).



Level 1: 4 (exact)
Level 2: 8 (exact)
Level 3: 9 (exact)
Level 4: 11 (exact)
Level 5:
Level 6:
Level 7:
Level 8:
Level 9: 17 (could be as low as 14)

It looks like for high levels the solution with all 1s is also the lower limit (my solution for 9 was done before I'd done this a lot - note that it's not true for Level 1), but that's partly because so far I've been lucky to have two convenient products close to the selected primes (12 and 18 for level 2, 18 and 24 for level 3, 32 and 36 for level 4). Level 5 has 48 and 54 so it should work, but level 6 only has 60 (which isn't that good), level 7 has only 72, level 8 is 96, and level 9 is 108. These aren't the only convenient products (100 is good for level 9 too), but it does suggest that the solution with only 1s may not be the 'hardest' one, so more dice can be required.

[Sith_Happens]

our admittedly rusty math and statistics knowledge.

I don't see where statistics comes into it. Meschlum is currently determining the minimums for guaranteed success. If he were looking for the minimums that provide, say, a 99.9% chance of success, that would involve statistics. I also suspect that most of those minimums are at least several ranks lower.

[grarrrg]

Minimum all 1's
For the sake of space and sanity, a "3" is actually "1+1+1"
Level 1: 3 (3 = 3)
Level 2: 8 (11 = 3*3+2)
Level 3: 9 (19 = 3*3*2+1)
Level 4: 11 (37 = 3*3*2*2+1)
Level 5: 12 (53 = 3*3*3*2-1)
Level 6: 13 (61 = 5*3*2*2+1 (holy cow! a FIVE!))
Level 7: 13 (73 = 3*3*2*2*2+1)
Level 8: 14 (83 = 3*3*3*3+2)
Level 9: 14 (107 = 3*3*3*2*2-1)

I agree that level 6 might be tricky, mainly due to me having to resort to a dirty, icky 5 to get to it.
Also, level 8 might be easier than you think. It winds up having the largest "spread" of all of them (i.e. level 1 ranges from 3 -to- 7, has a 5 point spread, level 8 ranges from 83 -to- 97, has a 15 point spread, the rest are all 7, 9 or 11). So while there is no 'easy' jump-in point, there are more potential jump-in points.

"Hail Mary" minimum (absolute lowest needed for a _possible_ success, however unlikely)
Level 1: 1 (3 = 3)
Level 2: 2 (11 = 6+5)
Level 3: 3 (19 = 6*3+1)
Level 4: 3 (37 = 6*6+1)
Level 5: 4 (43 = 6*6+6+1)
Level 6: 4 (61 = 6*5*2+1)
Level 7: 4 (73 = 6*6*2+1)
Level 8: 4 (89 = 6*5*3-1)
Level 9: 4 (107 = 6*6*3-1)
A couple of these go down a notch if using d8's with Calculating Mind (level 6 drops to 3 > 8*8-3)

[Khosan]

What you can also do to make it easier is break it up into sets of...2-3 or so.

At 2 or 3 dice, there are 21 or 56 possible combinations respectively you can get, which is pretty easy to solve for. You'd just need to list out every possible outcome from any possible combination of dice (and the equation that gets you that, but I figure the DM would stop asking by the umpteenth time). It should work pretty well, saving a lot of time and pretty much allowing you turn a great big pile of rolls into something meaningful.

I guess as an example:

17d6 - Apparently that doesn't work any more

I'll try to solve for a 9th level spell. Might take me a bit. I don't have a table of that information in front of me and it's 2AM so I'm not making it.

EDIT: http://invisiblecastle.com/roller/view/4592457/

So 3 1s, 5 2s, 3 3s, 2 4s, 1 5 and 3 6s. Looking to manage 101, 103 or 107.

1, 2, 2 gives me any value between 0 and 6, it's like a wildcard (and, I assume, a pile of 3-number combinations do the same), and I'll take the 6. Multiply by 6, multiply by 3 and we're up at 108. Subtract 1 and then clear the rest. Done.

EDIT: Actually, better would be to find those 'wildcard' number combinations that can solve into anything. They pretty much free you up to do whatever. After that, all you need is a free odd number to get you to the prime.

[Hecuba]

Some thoughts on doing this in a timely fashion.


3 1s, 5 2s, 3 3s, 2 4s, 1 5 and 3 6s. Looking to manage 101, 103 or 107 (borrowing you're rolls as an example).

Subtract a small number from the top prime target to get something that factors well, then negate the remaining dice. If you can't negate, you can likely just subtract & move down 1 target.

107-2 is 105. Thus

107=105+2=2+7*3*5=2+(3+4)*3*5

That leaves us

6-6+2*3-6+4-2-2+2/1-1-1=0

It won't always work, but it often will. And it will be much faster than working from the dice (since you can know the convenient decompositions of the target primes in advance).

[meschlum]

While there are benefits to processing a few dice at a time, there are no hard and fast rules for the number of dice you need to get a given value (it's the point of the exercise), so you're likely to wildly overestimate the number of dice needed if you break down the requirements that way. Using pairs to get 1 and 0 is the really critical part, as is knowing that four or more dice can always cancel out.

And you can figure out decompositions just by looking at the calculations I perform. Not always optimal, but guaranteed to work (other solutions exist too, usually) - the overall method is "Get the highest dice, see how you can combine them to come close, then tinker with the rest".



If you have two 6s and a 4, (6 + 6) * 4 - 1 works, using the known dice and a pair. The rest cancels out.

If you have two 6s and a 2, (6 + 2) * 6 - 1 works, with the rest canceling out.

If you have two 6s and a 5, (5 + 2) * 6 + 1 works, with the rest canceling out.

If you have two 6s and a 3, (6 + 3) * 6 - 1 works, with the rest canceling out.

So if you have two 6s, all dice are 1s or 6s. With three or more 1s, you have (6 + 1 + 1) * 6 - 1, using five dice and the remaining seven dice cancel out. Therefore, there are at most two 1s and at least ten 6s. If the 1s are replaced with pairs of 6s in the equation, eight 6s are used, which is possible since there are at least ten. This leaves four or more dice, so the rest cancel out.

Therefore, there is at most a single 6.

If there are two 4s and a 3, 4 * 4 * 3 - 1 is a solution, using a pair and the rest cancels out.

If there are three 4s, 4 * 4 * (4 - 1) - 1 is a solution, using two pairs (possible with nine dice), and the rest cancels out.

If there are two 4s and a 2, 4 * 4 * (2 + 1) - 1 is a solution, with the rest canceling out.

If there are two 4s and a 6, 6 * (4 + 4) - 1 is a solution using one pair, and the rest cancels.

If there are two 4s and a 5, (5 + 1) * (4 + 4) - 1 is a solution using two pairs, and the rest cancels.

So if there are two 4s, the other dice are 1s, leading to ten or more 1s. 4 * 4 * (1 + 1 + 1) - 1 is a solution using six dice in this case.

Therefore, there is at most a single 4.

If there are two 3s and a 6, 3 * 3 * 6 - 1 is a solution using one pair, and the rest cancels out.

If there are two 3s and a 5, 3 * 3 * (5 + 1) - 1 is a solution using two pairs, and the rest cancels out.

If there are two 3s and a 4, 3 * 4 * (3 + 1) - 1 is a solution using two pairs, and the rest cancels out.

So if there are two or more 3s, all the other dice are 1s, 2s, or 3s.

From 3 * 3 * 3 * 2 - 1, using 3s and 2s as well as a pair (as there are 8 dice left after the 3s and 2s), we know the other dice cannot have 3s and 2s together.

If there is at least one extra 3, all the dice are 3s and 1s, 3 * 3 * 3 * (1 + 1) - 1 means there are two 1s at most, so ten or more 3s. (3 + 3) * 3 * 3 - 3/3 uses six dice, so the rest cancel out.

So there are two 3s at most, and the other dice are 1s and 2s. 3 * 3 * (2 + 2 + 2) - 2/2 is a seven die solution (canceling the rest), so there are four 2s at most, hence at least six 1s. 3 * 3 * (1 + 1 + 1) * (1 + 1) - 1 uses six 1s, and so is possible, using eight dice. The remaining four cancel out.

Thus there is at most a single 3.

If there is a 6, a 2, and a 5, then 6 * (2 + 5) + 1 is a solution. Thus, if there is a 6, there are either 1s and 5s or 1s and 2s besides the other wild die.

If there is a 6 and a 5, 6 * (5 + 1 + 1) + 1 is a solution, so there are two 1s at most. This means that there are at least eight 5s (6, 5, 1, 1, and a wild die are set, the remaining are 5s). (6 + 5) * 5 - (5 + 5) / 5 is a solution, using six dice and five 5s, so it works and the rest cancel out. Thus if there is a 6, all the other (non wild) dice are 1s or 2s.

6 * 2 * 2 * 2 - 2/2 is a solution with five 2s and six dice in all, so there are at most four 2s and at least six 1s (plus the wild die).

6 * (1 + 1 + 1 + 1) * (1 + 1) - 1 is a solution with six 6s and a pair, for nine dice in all. After using a 6 and six 1s, there are four non-wild dice left, which are all 1s or 2s, so there is a pair among them. The three or more dice left are 1s, 2s, and a wild die (which is not 6). If the wild die is 1 or 2, there is a pair and the remainder cancels out. If the wild die is a 4, it can replace four of the 1s, which leaves a pair of 1s to cancel the rest. If the wild die is a 3, it an replace three of the 1s, with the same result. Thus, there is always a solution when one of the wild dice is a 6, so the wild dice are either 3 or 4.

We therefore have ten dice that are 1, 2, or 5 and two wild ones that can be anything but 6 (but with a maximum of one 3 and one 4).

If there is a 3 and a 4, 3 * 4 * (2 + 2) - 2/2 means that there are at most three 2s. Replacing the 2s with 1 + 1 (and the 2/2 with a single 1) means that there are at most four 1s. Therefore, there are at least three 5s. 3 * 4 * (5 - 5/5) - 1 is a solution using one pair, for a total of seven dice. The remaining dice can cancel, so there cannot be a 3 and a 4, meaning there is a single 'wild' die which can be 3 or 4.

With three or more 5s, 5 * (5 + 5 - 1) - 2 is a solution, so there must be only 5s and 2s or only 5s and 1s.

5 * (5 + 5 - 5/5) - (5 + 5) / 5 is a solution using eight 5s, so there are at most seven 5s, and there are 2s or 1s. The remaining four or more dice cancel out.

With three or more 5s and three or more 2s, 5 * (5 + 5 - 2/2) - 2 is a solution, so if there are three or more 5s, there are two 2s at most. In this case, as there are eleven non wild dice, this means there are at least nine 5s, wich is enough for the solution above.

Similar reasoning can be applied with three or more 5s and three or more 1s, using 5 * (5 + 5 - 1) - 1 - 1.

Thus there are two 5s at most.

We have (5 + 2) * (5 + 2/2) + 2/2 as a solution, so if there are two 5s, there are at most four 2s. This means there are at least five 1s. Since (5 + 1 + 1) * (5 + 1) + 1 is a solution using four 1s, there is always a solution with two 5s.

Therefore, there is a single 5 at most.

If there is a 4 and a 5 (a 3 and a 5), then (4 + 5) * (1 + 1) * (1 + 1 + 1) - 1 (or (3 + 5) * (1 + 1 + 1) * (1 + 1) - 1)) is a solution using six 1s. Thus there are at most five 1s, and at least five 2s.

With a 4 and a 5, we have (4 + 2) * (5 + 2) + 2/2 as a six die solution with four 2s, so it is possibile, and there cannot be a 4 and a 5.

With a 3 and a 5, we have (3 + 5) * (2 + 2 + 2) - 2/2 as a seven die solution with five 2s, so it is possible and there cannot be a 3 and a 5.

Thus, there is a single wild die that can be 3, 4, or 5 and the other eleven dice are 1s and 2s.

From (5 + 2) * (2 + 2 + 2) + 2/2, we see that there are at most five 2s if there is a 5, hence at least six 1s.

A 5 and two or more 2s (and six or more 1s) is solved by (5 + 1 + 1) * (2 + 2 + 1 + 1) + 1, using eight dice, so the rest cancels out. Hence if there is a 5, there is at most a single 2 and the other dice are 1s (at least ten).

With 5 * (1 + 1 + 1) * (1 + 1 + 1) - 1 - 1, we have a nine die solution using eight 1s. This leaves at least two 1s, which cancel the rest via 1-1.

So the wild die cannot be 5.

From 4 * 2 * (2 + 2 + 2) - 2/2, we see that there are at most five 2s if there is a 4, so at least six 1s.

A 4 and two or more 2s (and six or more 1s) is solved by 4 * 2 * (2 + 1 + 1 + 1 + 1) - 1, using eight dice. Thus there is a single 2 at most and ten or more 1s if there is a 4.

With 4 * (1 + 1 + 1) * (1 + 1 + 1 + 1) - 1, we have a nine die solution using eight 1s, so there is a pair of 1s left to cancel the remainder.

So the wild die cannot be 4.

From 3 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1) - 1, we get a ten die solution using nine 1s. The remaining dice are 1s and 2s, and will cancel out if there is a pair. If there is no pair, multiplying by 2-1 will cancel the rest. Thus there are eight 1s at most, and at least three 2s.

With 3 * 2 * 2 * 2 * 2 - 2/2, we get a solution using six 2s (and seven dice in all). Therefore, there are at most five 2s, and at least six 1s.

We get 3 * 2 * 2 * 2 * (1 + 1) - 1 using three 2s and three 1s, which satisfies the conditions above. Therefore there is always a solution when there is a 3.

So there is no wild die, and all dice must be 1s or 2s.

2 * 2 * 2 * 2 * (2 + 2/2) - 2/2 is a nine die solution. The remainder or three or more dice is 1s and 2s, so there is at least a pair to cancel it. Thus, there are at most eight 2s and at least four 1s.

With four or more 2s and at least four 1s, we get 2 * 2 * 2 * 2 * (1 + 1 + 1) - 1 is an eight die solution (the remaining four or more dice cancel out), so there must be three or less 2s (and nine or more 1s).

With two or more 2s and at least nine 1s, we get 2 * 2 * (1 + 1 + 1 + 1) * (1 + 1 + 1) - 1 as a ten die solution. The other two or more dice are 1s and 2s. If there is a pair, they cancel out. If there is none, then multiply by 2-1 to cancel the remaining two dice.

With one 2 and at least eleven 1s, we get 2 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) - 1 as an eleven die solution. The remainingoe or more dice are 1s, so they cancel via multiplication

With twelve 1s, there is a solution as well.



Level 1: 4 dice (exact)
Level 2: 8 dice (exact)
Level 3: 9 dice (exact)
Level 4: 11 dice (exact)
Level 5: 12 dice (exact)
Level 6:
Level 7:
Level 8:
Level 9: 17 dice (may be as few as 14)

[meschlum]

So... time to see how hard things get for Bards and the like.



With three 6s, you have 6 * (6 + 6 - 1) + 1, using two pairs (allowed as you have ten 'free' dice) and seven dice in all, so the remainder cancels out. Thus you have two 6s at most.

With a 6 and a 5, you have 6 * 5 * (1 + 1) - 1, using three pairs (which your eleven 'free' dice allow) and eight dice in all, so the remainder cancels out.

With two 6s and a 4, you have 6 * (6 + 4) + 1, using one pair and five dice in all.

With two 6s and a 3, you have 6 * (6 + 3 + 1) + 1, using two pairs (possible as there are ten 'free' dice) and seven dice, so it cancels out.

So if you have two 6s, the other dice are only 1s and 2s.

From 6 * (6 - 2/2) * 2 + 2/2, using seven dice (so the rest cancels), we see that there are four 2s at most, and at least seven 1s.

With 6 * (6 - 1) * (1 + 1) + 1 using six dice and four 1s, we see that all cases with wo 6s are valid (as are all cases with one 6 and a 5).

If there are three 5s, there are no 6s, and all the dice are 1s to 5s, so the remaining ten dice include at least three pairs. (5 + 1) * (5 + 5 + 1) + 1 is a solution using nine dice, and the rest cancels out as there are thirteen or more dice in all.

So there are two 5s at most.

If there are two 5s and a 4, all the ten or more other dice are 1s to 4s, allowing three pairs. (5 + 5) * (4 + 1 + 1) + 1 is a solution using all three pairs and nine dice in all. The remaining four or more dice cancel out.

If there are two 5s and a 3, he same reasoning applies, using (5 + 5) * 3 * (1 + 1) + 1 instead.

So if there are two 5s, all other dice must be 1s or 2s.

We have 5 * 2 * 2 * (2 + 2/2) + 2/2 as an eight die solution (the rest cancels), so there are six 2s at most, and at least six 1s.

We have 5 * 2 * (1 + 1) * (1 + 1 + 1) + 1 as an eight de solution using six 1s and a single 2, so there cannot be a 2 if there is a 5.

We have 5 * (1 + 1) * (1 + 1) * (1 + 1 + 1) + 1 as a nine die solution, with the rest canceling out (and all equal to 1 anyway).

So there is one 5 at most, and a single 'wild' die that can be equal to 5 or 6.

If there are three 4s, we have 4 * (4 - 1) * (4 + 1) + 1 as a solution, using nine dice. With nine 'free' dice (and a wild die) after setting three 4s, all in the 1 to 4 range, we have at least three pairs, so the solution is possible.

Therefore, there are two 4s at most.

Two 4s and a 6 give 4 * (4 - 1) * (6 - 1) + 1, using three pairs and nine dice. The remainder cancels out, and there are at least three pairs in ten dice ranging from 1 to 4, so it's possible.

Two 4s and a 5 give 4 * (4 - 1)* 5 + 1, using two pairs and seven dice, so if there are two 4s the other dice cannot be 5

Two 4s and a 3 give 4 * (4 + 1)* 3 + 1, using two pairs and seven dice, so if there are two 4s the other dice cannot be 3

So if there are two 4s, the other dice must be 1s or 2s.

We have 4 * (2 * 2 * 2 * 2 - 2/2) + 2/2 as a nine die solution with eight 2s, so there are at most seven 2s and at least five 1s.

We have 4 * (2 * 2 * 2 * (1 + 1) - 1) + 1 as an eight die solution with four 1s, so there are at most two 2s and at least ten 1s.

We have 4 * (1 + 1 + 1 + 1 + 1) * (1 + 1 + 1) + 1 as a ten die solution with nine 1s, and the remaining three or more dice are all 1s and 2s, so they include a pair, which cancels the rest.

Therefore, there is at most a single 4, and (apart from the 'wild' die) all other dice are 1s to 3s.

If there is a 4, a 3, and a 6, we have 4 * (6 - 1) * 3 + 1 as a solution using seven dice and two pairs (which is possible since there are ten dice in the 1 to 3 range). Therefore, with a 4 and a 6 (or a 4 and a 5, replacing the 6-1 with a 5), there cannot be any 3s.

If there is a 4, a 2, and a 6, we have 6 * 2 * (4 + 1) + 1 as a solution following the same criteria as above. Replacing the 6 with a 5 + 1 requires three pairs, which is possible since there are ten dice in the 1 to 2 range.

So if there is a 4 and a 6, or a 4 and a 5, the other dice must be 1s. 5 * 4 * (1 + 1 + 1) + 1 and (6 + 4) * (1 + 1 + 1) * (1 + 1) + 1 are solutions, meaning that 4 is on the 'wild' die only.

So we have twelve dice in the 1 to 3 range (guaranteeing five or more pairs), and one wild die which can be 1 to 6.

We have (3 + 3) * (3 * 3 + 1) + 1 as a solution using four 3s and two pairs (possible with eight dice in the 1 to 3 range), for eight dice in all. Thus there are three 3s at most.

With three 3s, we have nine or more 1s and 2s. (3 + 3) * (3 * 2 * 2 - 2) + 2/2 is a solution using five 2s and eight dice, so there are at most four 2s and five or more 1s. (3 + 3) * (3 * (1 + 1 + 1) + 1) + 1 is a solution using five 1s and eight dice, so there is always a solution if there are three 3s exactly.

With two 3s, we have ten or more 1s and 2s. (3 + 3) * (2 * 2 * 2 + 2) + 2/2 is a solution with six 2s and eight dice, so there are at most five 2s and five or more 1s.

We have (3 + 3) * (2 * 2 * (1 + 1) + 1 + 1) + 1 as a solution using five 1s, two 2s, and nine dice in all, so the rest cancels out and there is at most a single 2, so nine or more 1s.

We have (3 + 3) * ((1 + 1 + 1) * (1 + 1 + 1) + 1) + 1 as a solution using eight 1s and ten dice in all. The remaining dice include at least one 1, and the other two can be 1s, 2s, or the wild die. If there is a pair, the remainder cancels out, so the three dice must be 1, 2, and 3 to 6.

If the wild die is a 3, there are three 3s, and there is a solution.

If the wild die is a 4, we have (3 + 3) * (4 + 1) * (1 + 1) + 1 as a solution with seven dice and four 1s, so it is valid.

If the wild die is a 5, we have the same as the above, with 5 instead of 4+1

If the wild die is a 6 we have the same as the above, with 6-1 instead of 4+1

So the remaining three dice either cancel out or there is a solution using the wild die instead.

Therefore, there is a single 3 at most, and the other dice are 1s or 2s (besides the wild die).

With a 3 and a 4, we have 3 * 4 * (2 + 2 + 2/2) + 2/2, so there are five 2s at most (eight dice used, so the rest cancels), and at least six 1s. Six 1s can be used to replace the 2s in the solution exactly, so there is always a solution in this case.

With a 3 and a 6, we have 3 * (6 - 2/2) * (2 + 2) + 2/2, so there are five 2s at most, and at least six 1s. Again, the six 2s can be replaced with six 1s in the specified solution, so there is a solution in this case. If we replace the 6 with a 5 * 1 (or 5 * 2/2), the same formula applies.

Therefore, the wild die can be 3 to 6, and the twelve other dice are 1s and 2s.

From 3 * 2 * 2 * (2 + 2 + 2/2) + 2/2, we have a nine die solution using eight 2s, so there are seven 2s at most and at least five 1s.

With 3 * 2 * 2 * (2 + 1 + 1 + 1) + 1, we have an eight die solution using four 1s, so there are two 2s at most and at least ten 1s.

With 3 * (1 + 1 + 1 + 1 + 1) * (1 + 1 + 1 + 1) + 1, we have an eleven die solution using ten 1s, and the remaining two or more dice are either a pair (and cancel out) or 1 and 2, and we can multiply the result by 2-1.

So the wild die cannot be a 3.

With 4 * (2 + 2/2) * (2 + 2 + 2/2) + 2/2, we have a ten die solution using nine 2s. The three or more remaining dice are 1s or 2s, so include a pair and cancel out. Thus we have at most eight 2s and at least four 1s.

With 4 * (2 + 1) * (2 + 2 + 1) + 1, we have a seven die solution using three 2s and four 1s. Thus there are at most two 2s, and at least ten 1s.

With (4 + 1) * (1 + 1 + 1) * (1 + 1 + 1 + 1) + 1, we have a ten die solution using nine 1s. The remaining three or more dice are 1s or 2s, and so include a pair and cancel out.

So the wild die cannot be a 4.

With 5 * 2 * 2 * (2 + 2/2) + 2/2, we have an eight die solution using seven 2s. Thus there are at most six 2s, and at least six 1s.

With 5 * 2 * (1 + 1) * (1 + 1 + 1) + 1, we have an eight die solution using six 1s and a 2. Thus there are no 2s, and at least twelve 1s. The 2 can be replaced with two 1s, giving a nine die solution with eight 1s, so it's allowed.

So the wild die cannot be a 5.

With 6 * (2 + 2 + 2 + 2 + 2) + 2/2, we have an eight die solution using seven 2s. Thus there are at most six 2s, and at least six 1s.

With 6 * 2 * (1 + 1 + 1 + 1 + 1) + 1, we have an eight die solution using six 1s and a single 2. Thus there must be at least twelve 1s. We can replace the 2 with 1+1, for a nine die solution using a 6 and eight 1s.

So there is no wild die, and all dice must be 1s and 2s.

With 2 * 2 * (2 * 2 * 2 * 2 - 2/2) + 2/2, we have a ten die solution. The three or more remaining dice are 1s and 2s, so there is a pair and they cancel out. Thus there are at most nine 2s and at least four 1s.

With 2 * 2 * (2 * 2 * 2 * (1 + 1) - 1) + 1, we have a nine die solution using four 1s and five 2s. Thus there are at most four 2s, and at least nine 1s.

With 2 * 2 * (2 + 2 + 1) * (1 + 1 + 1) + 1, we have a nine die solution with four 2s, so there are three 2s at most (and at least ten 1s).

With 2 * 2 * (2 + 1) * (1 + 1 + 1 + 1 + 1) + 1, we have a ten die solution with three 2s exactly, and the remaining dice cancel out a they are all 1s. So there are two 2s at most.

With 2 * 2 * (1 + 1 + 1) * (1 + 1 + 1 + 1 + 1) + 1, we have an eleven die solution. Again, the 1s cancel out. The 2s can be replaced with 1+1, adding one die to the solution size. Since the remaining dice are all 1s, they cancel out anyway.

So 13 dice work out.


Level 1: 4 (exact)
Level 2: 8 (exact)
Level 3: 9 (exact)
Level 4: 11 (exact)
Level 5: 12 (exact)
Level 6: 13 (exact)
Level 7:
Level 8:
Level 9: 17 (could be as low as 14)

It's definitely looking like practice and the range of numbers we have mean that 14 dice are all you'll need at level 9.

Spontaneous casters (including level 6 limits like bards) are about 1 level behind at low levels, and then catch up or outgrow their limits (bards before sorcerers). Prepared casters are more like 2 levels behind up to level 12, and then catch up over the course of (probably) two levels. A rather traditional caster progression, granting lots of power at high levels for some inconvenience at low levels.

[meschlum]

There are numbers to crunch!



With two 6s, we have 6 * 6 * (1 + 1) + 1 as a solution, using three pairs (which is possible as there are eleven free dice) and eight dice, so the rest cancels out.

There is at most a single 6.

With three 4s, we have 4 * 4 * (4 + 1) - 1 as a solution, using two pairs (possible as there are ten free dice) and seven dice, so there are at most two 4s.

Two 4s and a 5 is a solution by the equation above (using a single pair)

Two 4s and a 6 is a solution by the equation above (with 6 - 1 instead of 4 + 1)

So if there are two 4s, all other dice are 1s to 3s.

With two 4s and two 3s, we have (4 + 4) * 3 * 3 + 1 using six dice, so there is one 3 at most.

With two 4s and one 3, all other dice are 1s and 2s. (4 + 4) * 3 * (2 + 2/2) + 2/2 uses eight dice, so there are at most four 2s and at least six 1s. (4 + 4) * 3 * (1 + 1 + 1) + 1 uses seven dice and four 1s, so there is always a solution with two 4s and a 3.

So if there are two 4s, the other dice are 1s and 2s.

We have 4 * 4 * (2 + 2 + 2/2) - 2/2 as an eight die solution, so there are at most five 2s, and at least six 1s. 4 * 4 * (1 + 1 + 1 + 1 + 1) - 1 uses six 1s and eight dice, so there is always a solution if there are two 4s.

With a 4 and a 6, we have 4 * 6 * 3 + 1 as a five die solution (including a pair), so there cannot be a 3. 4 * 6 * (2 + 1) + 1 is a seven die solution (including two pairs, possible as there are ten free dice), so there are no 2s.

With a 4 and a 6, the other eleven or more dice must be 1s or 5s. If there is a 5, there are four pairs, and we have 5 * 4 * (6 - 1 - 1) - 1 as a solution, using nine dice. The remaining four cancel out, so we have 4, 6, and 1s. 4 * 6 * (1 + 1 + 1) + 1 is a six die solution.

Therefore, there cannot be a 4 and a 6, so there is a single 'wild' die that can be 4 or 6, and the other dice are 1, 2, 3, or 5.

If there is one 3 and two 5s, we have 5 * (3 + 5) * (1 + 1) - 1 as a solution using three pairs (we have nine free non wild dice, so three pairs are possible as non wild dice have only four possible values) with nine dice in all, so the remaining four or more cancel out. So if there are two or more 5s, other non wild dice are 1, 2, or 5.

If there are four or more 5s, we have (5 * 5 - 5) * (5 - 1) - 1 as an eight die solution using two pairs. With eight non wild dice, two pairs are possible, so we have three 5s at most.

With exactly three 5s, we have (5 * 5 - 5) * (2 + 2) - 1 as a six die solution using a pair (there are eight non wild dice worth 1, 2, or 5, so one pair is possible). Therfore, there is a single 2 at most, and at least nine 1s. We have (5 * 5 - 5) * (1 + 1 + 1 + 1) - 1 as an eight die solution using five 1s, so there is always a solution with exactly three 5s.

With exactly two 5s, the remaining eleven or more dice are 1s and 2s (ensuring five pairs). We have (5 * 5 - 1) * (2 + 1) + 1 using three pairs and nine dice, so there cannot be a 2. Replacing the2 with two 1s, and the pairs with 1s as well, gives a seven die solution.

Therefore, there is at most a single 5, and other non wild dice are 1, 2, or 3.

If there is a 4 and a 5, we have 4 * 5 * (3 + 1) - 1 as a solution using two pairs and seven dice (there are ten free dice, so two pairs are possible). 4 * 5 * (2 + 2) - 1 is a solution as well, using six dice, so there is at most a single 2 and no 3s if there is a 4 and a 5. This means there are at least ten 1s, and 4 * 5 * (1 + 1 + 1 + 1) - 1 is a solution.

If there is a 6 and a 5, we have 6 * (5 + 1) * 3 + 1 as a solution using two pairs and seven dice again. Thus there cannot be any 3s in this case. 6 * (5 + 1) * (2 + 1) + 1 uses three pairs, drawn from ten free dice worth 1 or 2, which is possible. So with a 6 and a 5, the other dice are 1s, and we have 6 * (5 + 1) * (1 + 1 + 1) + 1 as a seven die solution.

Therefore, the wild die can be 4, 5, or 6 and the other dice are all 1s, 2s, and 3s.

If there are four 3s, 3 * 3* (3 * 3 - 1) + 1 uses eight dice and two pairs, so it is a solution. Therefore, there are at most three 3s.

If there are two or more 3s, 3 * 3 * 2 * 2 * 2 + 1 uses seven dice and one pair, so it is a solution. Therefore, if there are two or more 3s, there are at most two 2s. With three or fewer 3s and two 2s at most, this leaves seven 1s at least (plus the wild die).

With two 3s and seven 1s, we have 3 * 3 * (1 + 1) * (1 + 1) * (1 + 1) + 1 as a nine die solution, so the rest cancels out.

Therefore, there is a single 3 at most.

If there is a 3 and a 4, we have 3 * 4 * (2 + 2 + 2) + 1 as a seven die solution (one pair), so there are two 2s at most, and at least nine 1s. The solution can be reproduced by replacing 2 + 2 + 2 with (1 + 1) * (1 + 1 + 1), using six 1s in all, so there is always a solution.

If there is a 3 and a 5, we have 3 * (2 + 2) * (5 + 1) + 1 as an eight die solution (two pairs), so there is at most a single 2 and at least ten 1s. The solution can be reproduced by replacing 2 + 2 with 1 + 1 + 1 + 1, using six 1s in all, so there is always a solution.

If there is a 3 and a 6, we have 3 * 6 * (2 + 2) + 1 as a six die solution (one pair). As above, we can replace the 2s with 1s and get a solution as well.

Therefore, 3 can be present on the wild die only.

We have twelve dice that can be 1s or 2s, and one wild die which can be 1 to 6.

If there is a 6, we have 6 * (2 + 2) * (2 + 1) + 1 as an eight die solution using two pairs. There are therefore two 2s or less, and at least ten 1s. The three 2s in the solution can thus be replaced with six 1s for a nine die solution using eight 1s, and the rest cancels out.

If there is a 5, we have 5 * (2 + 2 + 2 + 2) - 1 as a seven die solution using one pair. There are therefore three 2s or less, and at least nine 1s. The four 2s in the solution can be replaced with eight 1s for a ten die solution using nine 1s. The remaining three dice are 1s or 2s, so there is a pair and they cancel out.

If there is a 4, we have 4 * (2 + 2 + 2) * (2 + 1) + 1 as a nine die solution using two pairs. There are therefore three 2s or less, and at least nine 1s. 4 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1) + 1 is ten die solution using nine 1s, and the remaining three dice are 1s or 2s, so there is a pair and they cancel out.

If there is a 3, we have 3 * (2 + 1) * 2 * 2 * 2 + 1 as a nine die solution using two pairs. There are therefore three 2s or less, and at least nine 1s.

If there is a 2, we have 3 * (1 + 1 + 1) * 2 * (1 + 1) * (1 + 1) + 1 as a ten die solution using eight 1s, and the remaining three dice include at least one pai as they are 1s and 2s.

If there is a 3, all the other dice are therefore 1s, and 3 * (1 + 1 + 1) * (1 + 1) * (1 + 1) * (1 + 1) + 1 is an eleven die solution. The remaining dice are 1s, so they cancel out.

Therefore, all the dice are 1s and 2s.

If there are five 2s, (2 + 1) * (2 + 1) * 2 * 2 * 2 + 1 uses eleven dice and three pairs, which is possible as there are eight dice which are equal to 1 or 2. The remaining two or more dice either have a pair and cancel out, or are 1 and 2, and multiplying by 2-1 cancels them. Therefore, there are at most four 2s, and at least nine 1s.

We can replace the 2s and pairs with 1s, to get (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1) * 2 * 2 + 1 with eleven dice and nine 1s. Again the remaining dice cancel out as they must be 1s or 2s, and there is at most a single 2 and at least twelve 1s.

2 * (1 + 1) * (1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) + 1 is a twelve die solution, and the remaining dice are 1s, so they cancel out.

There is a thirteen die solution using only 1s, so level 7 works as expected.



Level 1: 4
Level 2: 8
Level 3: 9
Level 4: 11
Level 5: 12
Level 6: 13
Level 7: 13
Level 8:
Level 9: 17 (probably 14)

An inverted approach, seeing the maximum number of 1s for each case, could work too, but it looks somewhat harder to extend to an arbitrary number of dice (if it works for X dice, you want to be sure that it also works for X + 1, X + 2, etc.)

[Snowbluff]

Nice work. This is great.

[meschlum]

Almost done. Time to try a two for one wall of spoilered text, shall we?



If you have a 4, 5, and 6, then you have eleven other dice, which guarantees three pairs. (4 - 1) * (5 + 1) * 6 - 1 is a nine die solution for Level 9, and 6 * 4 * (5 - 1) + 1 is a seven die solution for level 8. In both cases, the extra dice cancel out, so there is always a solution if you roll, 4, 5, and 6.

If you have two 4s and one 6, the level 8 solution requires even less dice, and 4 * (4 + 1) * (6 - 1) + 1 is a nine die solution for level 9.

If you have one 4 and two 6s, (6 + 6) * 4 * (1 + 1) + 1 is a nine die solution for level 8, and 6 * 6 * (4 - 1) - 1 is a seven die solution for level 9.

So if you have 4s and 6s, there is at most one of each, and no 5s.

The remaining twelve dice are 1s, 2s, and 3s, guaranteeing five pairs. 4 * 6 * (1 + 1 + 1 + 1) + 1 is a twelve die solution for level 8. If the remaining two or more dice are pairs or one apart (1, 2 or 2, 3), then they cancel out. If they are 1 and 3, then the (1 + 1 + 1 + 1) term can be replaced with (1 + 3), which leaves four pairs to cancel out the rest. If there is a 3, 6 * 3 * (4 + 1 + 1) - 1 is a nine die solution for level 9. Without 3s, there are twelve dice that are 1s and 2s, and (6 - 1) * (4 + 1) * 2 * 2 + 1 uses two 2s and three pairs (possible with ten dice that are 1s and 2s) for a total of ten dice. So there is at most a single 2, meaning there are at least eleven 1s, and the solution above is possible with seven 1s, with a total of nine dice.

So there cannot be 4s and 6s together.

Two 5s and one 6 give (6 - 1) * 5 * (5 - 1) + 1, a nine die solution for level 9, and 6 * (5 - 1) * (5 - 1) + 1, a nine die solution for level 8.

Two 6s and one 5 give 6 * 6 * (5 - 1 - 1) - 1, a nine die solution for level 9, and 6 * 6 * 5 / (1 + 1) - 1 is a nine die solution for level 8.

So if you have 5s and 6s, there is at most one of each, and no 4s.

The remaining twelve dice are 1s, 2s, and 3s, guaranteeing five pairs. (5 + 1) * 6 * (1 + 1 + 1) - 1 is a twelve die solution for level 9, and the remaining dice cancel out unless they are 3 and 1, in which case the (1 + 1 + 1) can be replaced by 3, leaving three pairs to cancel the rest. 5 * 6 * (1 + 1 + 1) - 1 is a ten die solution for level 8, and the rest cancels out.

So there cannot be 5s and 6s together, meaning that if there are one or more 6s, the other dice are 1s, 2s, and 3s.

Two 4s and one 5 give 4 * (4 + 1) * 5 +1 for level 9, and 4 * 4 *(5 + 1) + 1 for level 8.

Two 5s and one 4 give 4 * 5 * 5 + 1 for level 9, and 4 * (5 - 1) * (5 + 1) + 1 for level 9, a nine die solution.

So if you have 4s and 5s, there is at most one of each, and no 6s.

The remaining twelve dice are 1s, 2s, and 3s, guaranteeing five pairs. If there is a 3 and a 2 (so ten free dice, and four pairs), then 4 * 5 * (3 + 2) + 1 is a six die solution for level 9, and 4 * 2 * (5 + 1) * (3 - 1) + 1 is a solution for level 8 that uses three pairs and ten dice, and the other four cancel out. So the remaining twelve dice are 1s and 2s or 1s and 3s (and there are at least five pairs). If there is a 3, then 4 * 5 * (3 + 1 + 1) + 1 is a nine die solution for level 9, and 4 * (3 + 1) * (5 + 1) + 1 is a nine die solution for level 8. Thus the remaining twelve dice are 1s and 2s. If there is a 2, then 4 * 5 * (2 + 1 + 1 + 1) + 1 uses eleven dice as a solution for level 9, and the remaining three dice include a pair and so cancel out. 4 * 2 * (1 + 1) * (5 + 1) + 1 is an eleven die solution for level 8, and therefore works, using the same reasoning. Thus we have 4, 5, and twelve 1s. Replacing the 2s in the formula for using a 2, and replacing the pairs with 1s, gives eight dice for level 9 and level 8 solutions.

So there cannot be 4s and 5s together, meaning that we need to review cases with 1s, 2s, 3s, and 4s, cases with 1s, 2s, 3s, and 5s, and cases with 1s, 2s, 3s, and 6s.

If there are two 6s, there are twelve other dice, so at least four pairs, and we have 6 * 6 * (1 + 1 + 1) - 1 as a ten die solution for level 9, so there is at most a single 6. With level 8, we use 6 * (6 + 1) * (1 + 1) - 1 to get the same result.

If there are two 5s and a 3, we have 5 * 5 * (3 + 1) + 1 as a seven die solution for level 9, and 5 * (5 + 1) * 3 - 1 for level 8. So if there are two or more 5s, there are no 3s. Two 5s and a 2 gives 5 * 5 * (2 + 1 + 1) + 1 for level 9, and 5 * (5 + 1) * (2 + 1) - 1 for level 8, both as nine die solutions using three pairs. So if there are two or more 5s, there are only 5s and 1s, which means there are at least five pairs among the remaining twelve dice. Replacing the 2s in the previous equations with 1+1 (two pairs) gives twelve die solutions using five pairs. If the remaing two dice are a pair, they cancel out. Otherwise, we have a 5 and a 1, so there are at least three 5s and one 1 (and four pairs). In this case, 5 * 5 * (5 - 1) + 1 is a six die solution for level 9, and (5 + 5 - 1) * 5 * (1 + 1) - 1 uses ten dice and three pairs for level 8.

Therefore, there is at most a single 5.

With two 4s, there are twelve other dice and at least four pairs, so 4 * 4 * (1 + 1 + 1) + 1 is a ten die solution for level 8.

If there are two 4s and one 3 there are eleven other dice (at least four pairs), and we have 4 * 3 * (4 * (1 + 1) + 1) - 1 as an eleven die solution for level 9. The remaining dice cancel out if there are pairs, so they are three values from 1, 2, 3, 4. If there is a 2 remaining, it can replace the (1 + 1) in the equation, leaving two pairs to cancel everything. So the remaining dice are 1, 3, 4, and 4-3-1 is zero. Therefore, there are no 3s for level 9 if there are two 4s.

If there are three 4s, we have (4 + 1) * (4 + 1) * 4 + 1 as a nine die solution for level 9, so there are two 4s at most.

If there are two 4s and two 2s, we have (4 + 1) * (4 + 1) * 2 * 2 + 1 as a ten die solution for level 9, so there is one 2 at most, and there are at least eleven 1s. These can replace the 2s and pairs in the solution, for a nine die solution using seven 1s, so there is always a solution if there are two or more 4s.

Therefore, there is a single 'wild' die that can be equal to 4, 5, or 6, and the other thirteen dice are all 1s, 2s, or 3s.

3 * 3 * 3 * (3 + 1) - 1 is a solution for level 9 with four 3s and two pairs, so eight dice in all. Therefore, there are at most three 3s for level 9. 3 * 3 * (3 * 3 + 1) - 1 is a solution for level 8 with the same number of 3s, so the same restriction applies.

With three 3s, there are eleven other dice (with three possible values), which allows at least four pairs, and we have 3 * 3 * 3 * 4 - 1, 3 * 3 * 3 * (5 - 1) - 1 and (3 + 3) * 3 * 6 - 1 as six and eight die solutions for level 9 if there is a 4, 5, or 6, so for level 9 all dice must be 1s, 2s, and 3s if there are three 3s. Level 8 can be dealt with as 3 * (3 * 3 - 1) * 4 + 1, 3 * (3 + 5) * (3 + 1) + 1, and 3 * (3 + 6 - 1) * (3 + 1) + 1 for eight and ten die solutions. Therefore, if there are three 3s, the other dice must be 1s and 2s. 3 * 3 * 3 * 2 * (1 + 1) - 1 uses three pairs and ten dice for level 9, so level 9 has only 1s - and replacing the 2 and pairs with 1s gives a solution. (3 * 3 + 1) * 3 * (2 + 1) - 1 uses three pairs and ten dice for level 8, so the same reasoning applies.

There are therefore two 3s at most.

With two 3s, if the wild die is a 4, 5, or 6, we have eleven dice that are 1s or 2s, so five pairs. 3 * 3 * 4 * (1 + 1 + 1) - 1 is an eleven die solution for level 9, and the remaining three dice are 1s or 2s, so there is a pair and they cancel out. (3 + 1) * 3 * 4 * (1 + 1) + 1 is also an eleven die solution for level 8, so it works as well. 3 * 3 * (5 + 1) * (1 + 1) - 1 is another eleven die solution for level 9 and 3 * (3 * 5 + 1) * (1 + 1) + 1 is an eleven die solution for level 8. 3 * 3 * 6 * (1 + 1) - 1 is a nine die solution for level 9, and 6 * (3 + 3 + 1) * (1 + 1) - 1 is an eleven die solution for level 8.

So if there are two 3s, the twelve or more other dice are 1s and 2s, for at least five pairs (and two or more remaining dice will cancel out).

We have 3 * 3 * (2 + 1) * (2 + 2) - 1 as a nine die solution for level 9, so there are at most two 2s if there are two 3s. Level 8 is done via (3 + 1) * (2 + 2) * 2 * 3 + 1, for the same criterion. With two 3s and two 2s at most, we have at least ten 1s, and the 2s and pairs in the earlier solutions can be replaced by these 1s, eight in all. This gives ten die solutions, so there is always a solution if there are two 3s.

With one 3 and the wild die at 4, 5, or 6, we have twelve dice that are 1s or 2s, so five pairs. 3 * 4 * 2 * 2 * (1 + 1) + 1 is a ten die solution for level 8, and 3 * 4 * (2 + 1) * (2 + 1) - 1 is its level 9 analog. Replacing the 4 with 5-1 gives a twelve die solution in either case, and the remaining two dice are 1s and 2s, so they cancel out (either a pair or 2-1). (3 + 1) * 6 * 2 * 2 + 1 is an eight die solution for level 8, and (3 + 6) * 2 * 2 * (1 + 1 + 1) - 1 is a twelve die solution for level 9 (the remaining two dice cancel out as they are 1s and 2s).

So the wild die includes 3, 4, 5, and 6, and the other thirteen dice are all 1s or 2s (six pairs).

2 * 2 * 2 * 2 * 2 * (2 + 1) + 1 is a ten die solution for level 8, using two pairs. Thus level 8 has five 2s at most, and at least eight 1s.

2 * 2 * 2 * (1 + 1) * (1 + 1) * (1 + 1 + 1) + 1 is an eleven die solution for level 8, using eight 1s. The remaining dice are 1s, 2s, and the wild die, so they cancel out unless they are 1, 2, and the wild die. If the wild die is 3, 3-2-1 cancels out. With 4, muiplying by 4-2-1 works. With 5, we can replace (1 + 1) * (1 + 1 + 1) with 5 + 1, so there are multiple 1s left and they cancel out. The same is done if the wild die is 6.

Therefore, there are at most two 2s, and at least eleven 1s. We can replace one of the 2s in the solution above with (1 + 1), for a twelve die solution using two 2s and ten 1s, and the remaining dice are a 1 and the wild die. If it is 1 or 2, it cancels out directly. If it is 3, it can replace the (1 + 1 + 1), and the rest cancels out. Similar replacements are possible for the other wild die values.

So there is at most one 2, and twelve 1s. Another 2 can be replaced, leaving a thirteen die solution with one 2 and the wild die. Again, replacements are possible for all the wild die values.

There is a solution for level 8 with fourteen 1s.

For level 9, we have (2 + 2 + 2) * (2 + 2 + 2) * (2 + 1) - 1 as an eleven die solution using two pairs. A wild die of 3, 4, or 6 can be replaced to cancel the remainder. If the remaining dice are 1, 2, and 5, we can replace three 2s with 1 + 5, which gives a pair of 2s to cancel the rest. So there are at most six 2s, and at least seven 1s.

With seven 1s, we can use 2 * (2 + 2 + 2) * (1 + 1 + 1) * (1 + 1 + 1) - 1, as an eleven die solution using seven 1s. Again, a wild die of 3, 4, or 6 can be replaced, and 2 + 2 + 2 can be replaced by 1+ 5. So there are at most three 2s, and at least ten 1s.

With ten 1s, we can use 2 * 2 * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) - 1, a twelve die solution using ten 1s. The remaining dice are 1s, 2s, or the wild values. 1s, 2s, and pairs cancel out. Again, 3, 4, and 6 can be replaced, so the options for the remaining dice are 1, 5 and 2, 5. The 5 can replace 2 * (1 + 1 + 1) with 5 + 1, freeing a 2 and two 1s, so there is a pair to cancel the remainder. Thus, there is a single 2 at most, and at least twelve 1s.

This gives 2 * (1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) - 1, a thirteen die solution. The remaining die is the wild die. A value of 1 cancels out, 2 replaces (1 + 1) and the pair of 1s cancel out, 3, 4, and 6 can also be cancelled, and 5 can be replaced as well.

Thus there are no 2s in the solution, and there is a fourteen die solution for level 9.



Level 1: 4 dice
Level 2: 8 dice
Level 3: 9 dice
Level 4: 11 dice
Level 5: 12 dice
Level 6: 13 dice
Level 7: 13 dice
Level 8: 14 dice
Level 9: 14 dice

Using more than one prime as the target is critical to getting the results this low. Beyond that, the general outcome is that you need lots of dice only if you roll a lot of 1s and 2s. Get a few 4s, 5s, and 6s, and you're pretty much set with (possibly much) lower dice pools.

Sadly, this ends my math frenzy. Maybe Pathfinder will come up with some other silly numeric manipulation?

[Serafina]

Congratulations, you just made Sacred Geometry usable - a player can just point to your results, treat the minimum number of dice as a minimum rank requirement and just skip actually doing all the math since there is a proven solution anyway.

Which is a good thing since it will prevent huge slowdowns, and a bad thing since the feat is still overpowered. Oh Paizo :smallsigh:

If you want more math, you could analyze the follow-up feat that allows using D8s :smalltongue:

[grarrrg]

Level 1: 4 dice
Level 2: 8 dice
Level 3: 9 dice
Level 4: 11 dice
Level 5: 12 dice
Level 6: 13 dice
Level 7: 13 dice
Level 8: 14 dice
Level 9: 14 dice

This exactly matches the "all 1's" results.
What a bunch of wasted time! You could have just gone with the all 1's numbers and been done! :smallwink:

But seriously.
Low levels the feat isn't that good, mainly due to lack of options/spell-levels "Woo, I can get a Free +1 Meta on a Cantrip!"
But once you hit level level 13+ everything auto-succeeds making you grossly uber-powerful. :smallsigh:

Good job Paizo...good job...you just made Casters better. Again. Because we all know they needed help. Especially at high levels.

[Kantolin]

Well, also remember: This is just the 100% success rate.

You can get a 'you are extremely unlikely to fail' success rate from just having six or seven dice. Even with five you're unlikely to fail, especially a +1.

[Snowbluff]

Congratulations, you just made Sacred Geometry usable - a player can just point to your results, treat the minimum number of dice as a minimum rank requirement and just skip actually doing all the math since there is a proven solution anyway.

Which is a good thing since it will prevent huge slowdowns, and a bad thing since the feat is still overpowered. Oh Paizo :smallsigh:


Yeah, you lose the fun and slowdown for power and faster pacing. Which is a good trade off. Like I said earlier, I wish their was a way to make this feat functional while still requiring a puzzle.

I suggest Echoing and Quicken as MM for this. Good action economy and longevity for your caster. However, you can take this multiple times for greater effect.

[grarrrg]

so long as you have four or more dice, you can generate a zero

Was thinking about the "4+ dice go bye-bye" thing today.
It's actually _3_+ dice go bye-bye.

While you do need 4 (or more) to guarantee a 0 or a 1, 3 dice will guaranteed get you a 0, a 1, or combine to make any fourth number you add.
Example:
4, 5, 6 will not get a 0 or a 1, BUT
4, 5, 6, 1 > (6+4)/5 - 1 = 1
4, 5, 6, 2 > 5 - 6*2/4 = 2
4, 5, 6, 3 > 6*4/3 - 5 = 3
4, 5, 6, 4 > (6*4 - 4)/5 = 4
4, 5, 6, 5 > (6-4)*5 - 5 = 5
4, 5, 6, 6 > (6*5 - 6)/4 = 6

I've ran the numbers, and this works for all combinations of 3 dice (will post if needed, or if I get bored later).
If 3 dice do not reduce to a 0 or a 1, they WILL let you "convert" any other number into itself.
Start with all Unique 3-die combinations (56 unique), then remove any with a duplicate number, as these easily reduce to 0 (20 non-doubles). Finally, remove all the combinations that reduce to 0 or 1. This leaves 5 combinations:
1, 2, 5
1, 2, 6
3, 4, 5
3, 5, 6
4, 5, 6
We already proved 4 ,5, 6 in the example

1, 2, 5, 1 > 5-2-1-1 = 1
1, 2, 5, 2 > 5-2-2+1 = 2
1, 2, 5, 3 > 5-3+2-1 = 3
1, 2, 5, 4 > 5-4+2+1 = 4
1, 2, 5, 5 > (5*2-5)*1 = 5
1, 2, 5, 6 > (5+1)*2-6 = 6

1, 2, 6, 1 > 6/2-1-1 = 1
1, 2, 6, 2 > (6-2-2)*1 = 2
1, 2, 6, 3 > 6/3+2-1 = 3
1, 2, 6, 4 > (6-4+2)*1 = 4
1, 2, 6, 5 > (6-1)*2-5 = 5
1, 2, 6, 6 > (6*2-6)*1 = 6

3, 4, 5, 1 > 4+3-5-1 = 1
3, 4, 5, 2 > 5+3-4-2 = 2
3, 4, 5, 3 > 5*3-4*3 = 3
3, 4, 5, 4 > (5-3)*4-4 = 4
3, 4, 5, 5 > 5*4-5*3 = 5
3, 4, 5, 6 > 5+4+3-6 = 6

3, 5, 6, 1 > 5+3-6-1 = 1
3, 5, 6, 2 > 6+3-5-2 = 2
3, 5, 6, 3 > (5*3-6)/3 = 3
3, 5, 6, 4 > 6+5-4-3 = 4
3, 5, 6, 5 > 6*5/3-5 = 5
3, 5, 6, 6 > 6+6/(5-3) = 6

Apologizes if I typed a wrong symbol and missed it.

[meschlum]

This exactly matches the "all 1's" results.
What a bunch of wasted time! You could have just gone with the all 1's numbers and been done! :smallwink:

Not quite: at the dangerously overpowered levels, you are a bit behind: you only need 3 dice to succeed with "all 1s" for a level 1 spell.


But seriously.
Low levels the feat isn't that good, mainly due to lack of options/spell-levels "Woo, I can get a Free +1 Meta on a Cantrip!"
But once you hit level level 13+ everything auto-succeeds making you grossly uber-powerful. :smallsigh:

Good job Paizo...good job...you just made Casters better. Again. Because we all know they needed help. Especially at high levels.

Obviously, since it's harder to get auto-successes at low levels than high ones!


Nice trick on the 3-dice bit! My solutions are somewhat brute force and had a certain amount of learning involved to get the most effective approaches (plus I experimented a bit), but refinements are always nice.


In return, an insight into the workings of the d8 option: it's never better, since the player can always have all the d8s come up in the 1-6 range, so the solutions using d6s are lower bounds, and the d8 solutions could be worse.

For instance, with d2s, there is always a 3 die solution for level 1 - but it goes up to four dice with d6s.



First, let's see what we can cancel out. The answer is 4 dice to get 0 and 1.

If all the dice come up in the 1 to 6 range, you need four (or three) to cancel everything.

Otherwise...

Doubles are excluded, of course.

If one of the dice is a 1, none of the others can be adjacent. (e.g. 1, 4, 5)
If one of the dice is a 2, none of the others can be 2 apart. (e.g. 2, 5, 7)
If one of the dice is a 3, none of the others can be 3 apart. (e.g. 3, 4, 7)

At least one die must be 7 or 8

8 / 4 - 2 = 0 removes some options.

If there are two pairs of numbers that are adjacent, or two apart, then they cancel out (e.g. (7 - 5) - (4 - 2))

If the lowest die is a 5, the four dice are adjacent, so they cancel out.

If the lowest die is a 4, and the highest is a 7, there are two adjacent pairs. So if the lowest die is a 4, the highest is an 8.

4, 5, 6, 8: 4 + 8 - 5 - 6 = 1. If another die is added, there is a pair, two sets of adjacent numbers, or the lowest die is less than 4.
4, 6, 7, 8: 6 + 7 - 4 - 8 = 1. If another die is added, there is a pair, two sets of adjacent numbers, or the lowest die is less than 4.

If the lowest die is a 3, none of the other dice can be 3 apart (4, 7 or 5, 8).
3, 4, 6, 8: 3 + 8 - 4 - 6 = 1. If another die is added, there is a pair, two sets of adjacent numbers, or the lowest die is less than 3.
3, 5, 6, 7: 5 + 6 - 3 - 7 = 1. If another die is added, there is a pair, two sets of adjacent numbers, or the lowest die is less than 3.

If the lowest die is a 2, none of the other dice can be 2 apart (3, 5; 4, 6; 5, 7; or 6, 8), and we also cannot have 4, 8.

2, 3, 4, 7: rule of 3 (4, 7)
2, 3, 4, 8: 2, 4, 8 present
2, 3, 5, 7: rule of 2 (5, 7)
2, 3, 5, 8: rule of 3 (5, 8)
2, 3, 6, 7: two adjacent pairs
2, 3, 6, 8: rule of 2 (6, 8)
2, 3, 7, 8: two adjacent pairs
2, 4, 5, 7: rule of 2 (5, 7)
2, 4, 5, 8: 2, 4, 8 present
2, 4, 6, 7: rule of 2 (4, 6)
2, 4, 6, 8: 2, 4, 8 present
2, 4, 7, 8: 2, 4, 8 present
2, 5, 6, 7: rule of 2 (5, 7)
2, 5, 6, 8: rule of 2 (6, 8)
2, 5, 7, 8: rule of 2 (5, 7)
2, 6, 7, 8: rule of 2 (6, 8)

So the lowest die must be a 1, limiting options further.

1, 2, 4, 7: 7 - 4 - 2 - 1 = 0
1, 2, 5, 8: 8 - 5 - 2 - 1 = 0
1, 3, 5, 7: (7 - 5) - (3 - 1) = 0
1, 3, 6, 8: (8 - 6) - (3 - 1) = 0
1, 4, 6, 8: 4 + 6 - 1 - 8 = 1. If another die is added, there is a pair, adjacent numbers, or 1, 2, 4, 6, 8, and 2, 4, 8 cancels out.

Therefore, with 4 dice, it is always possible to cancel out the d8s.


Now let's see what fate awaits an arithmancer trying to use four d8s and a Level 1 spell.

There must be at least one 7 or one 8 (or some of both), as solutions with 6s and less are known.

If there are two 8s, two 7s, or a 7 and an 8, thy can be compressed to 1.

This leaves the following options:

1, 1: 1 + 1 + 1
1, 2: 1 + 2 * 1
1, 3: 1 + 3 + 1
1, 4: 1 + 4 * 1
1, 5: 1 + 5 + 1
1, 6: 1 + 6 * 1
1, 7: 1 + 7 - 1
1, 8: 1 * 7 * 1
2, 2: 2 + 2 + 1
2, 3: 2 + 3 * 1
2, 4: 2 + 4 + 1
2, 5: 2 + 5 * 1
2, 6: 2 + 6 - 1
2, 7: 7 - 2 * 1
2, 8: 8 - 2 + 1
3, 3: 3 + 3 + 1
3, 4: 3 + 4 * 1
3, 5: 3 + 5 - 1
3, 6: 6 - 3 * 1
3, 7: 7 - 3 + 1
3, 8: 8 - 3 * 1
4, 4: 4 + 4 - 1
4, 5: no solutions with 1, 4, 5
4, 6: 6 - 4 + 1
4, 7: 7 - 4 * 1
4, 8: 8 - 4 + 1
5, 5: no solutions with 1, 5, 5
5, 6: no solutions with 1, 5, 6
5, 7: 7 - 5 + 1
5, 8: 8 - 5 * 1
6, 6: no solutions with 1, 6, 6
6, 7: no solutions with 1, 6, 7
6, 8: 8 - 6 + 1
7, 8: no solutions with 1, 7, 8
8, 8: no solutions with 1, 8, 8

With two 8s, if either of the other dice is a 7, we have (8 - 8) * X + 7 as a solution, so the remaining case is four 8s, which is solved by (8 + 8 + 8) / 8.
With 7 and 8, we have 7, 7, 7, 8 or 7, 7, 8, 8 or 7, 8, 8, 8. All of these are solved by taking 0 * X + 7, where the 0 is created with a pair.
With 6 and 7, we have 6, 7, 8, 8 (solved as above), 6, 7, 7, 8 (solved by converting 7, 7 into a 1), and 6, 7, 7, 7 (solved as above).
With 6 and 6, we have 6, 6, 8, 8 (solved via 8 - 6 / (8 - 6)), 6, 6, 7, 8 (solved as above), and 6, 6, 7, 7 (solved as above).
With 5 and 6, we have 5, 6, 8, 8 (solved as above), 5, 6, 7, 8 (solved by converting 6, 7 into a 1), and 5, 6, 7, 7 (solved as above)
With 5 and 5, we have 5, 5, 8, 8 (solved as above), 5, 5, 7, 8 (solved as above), and 5, 5, 7, 7 (solved as above)
With 4 and 5, we have 4, 5, 8, 8 (solved as above), 4, 5, 7, 8 (solved by 4 * 5 - 7 - 8), and 4, 5, 7, 7 (solved as above)

So there is at most a single die that is 7 or 8, and the other three are between 1 and 6.

If the wild die is a 7 and the other three dice are equal to 0 or 1, there is a solution. The remaining cases are:

1, 2, 5, 7: 7 + 2 - 1 - 5
1, 2, 6, 7: 7 + 1 - 6/2
1, 3, 5, 7: 7 + 3 - 5 * 1
3, 4, 5, 7: 7 - (3 + 5) / 4
3, 5, 6, 7: 7 + 6 - 3 - 5
4, 5, 6, 7: 7 - (4 + 6) / 5

So there is always a solution if there is a 7.

The wild die must therefore be an 8, and the other dice are 1s to 6s.

If there is a 6, 4, or 2, there is a solution if the other two dice are equal to 1 or 3 (and 5 if the die is not a 4). Pairs give 1, so the remaining cases are:

1, 5: 8 - (5 - 4) * 1
1, 6: 8 + 4 - 6 + 1
2, 4: 8 - 4 + 6 / 2, other cases have a pair (2, 4, 4, 8 and 2, 2, 4, 8)
3, 5: 8 - 5 + 6 / 3, 5 * (8 - 3 - 4), 3 * (8 - 5 - 2)
4, 6: see solutions for 2, 4

So all the other dice must be odd, and their sum ranges from 3 to 15.

3: 8 - 3
5: 8 - 5
7: to be reviewed
9: to be reviewed
11: 11 - 8
13: 13 - 8
15: 15 - 8

The sums to be reviewed can be obtained as follows:

1, 1, 5: 8 - 5 + 1 - 1
1, 3, 3: 8 - 1 + 3 - 3
1, 3, 5: 8 + 3 - 5 - 1
3, 3, 3: 8 - 3 + 3 - 3

So there is always a solution if there is an 8.


So with d8s, we have 4 dice for Level 1. No improvement at the certainty level.

[Sith_Happens]

Now that the minimum ranks for 100% success seem to have all been found, how about the minimum ranks for 99% success? I think a 1/100 fizzle chance is a more than acceptable risk for free metamagic under most circumstances.

[OldTrees1]

Now that the minimum ranks for 100% success seem to have all been found, how about the minimum ranks for 99% success? I think a 1/100 fizzle chance is a more than acceptable risk for free metamagic under most circumstances.

Something close to this can be determined by checking the probability for (Minimum dice - 1 die)

[Sith_Happens]

Something close to this can be determined by checking the probability for (Minimum dice - 1 die)

Cool, maybe someone who knows how to do that (meschlum?:smalltongue:) can even make a table of success chances for assorted target spell level and ranks combinations.

[meschlum]

Eh. With Level 2 or more (where "all 1s" is the worst case scenario), it's plausible that all 1s is the only case involving that many dice. Of course, the odds of rolling all 1s go below 1% as soon as you hit three dice, so it doesn't help much. Plus, there may be cases with other combinations that get close to (or at) the limit, so it's not guaranteed.

The thing with any approach but 100% success is that you need to work out all the combinations, and that is painful.


Still, a low level bonus:

Level 1 fails with 1, 4, 5 (among others), which occurs with probability 1/36 > 1%. Therefore, to have a 99% chance of success, you need more than 3 dice (so 4 dice). Happy?

[Yanisa]

Now that the minimum ranks for 100% success seem to have all been found, how about the minimum ranks for 99% success? I think a 1/100 fizzle chance is a more than acceptable risk for free metamagic under most circumstances.


Based on a 1000 rolls (http://i.imgur.com/VglJXiQ.png) (Credits (http://www.giantitp.com/forums/showsinglepost.php?p=17841832&postcount=51)), the magic number seems to be 10. You only really have to be afraid to roll the same number a bunch of times, because you can easier reach even 9th level spells with 4 numbers, this gives you already 6 dice for wiggle room, which is most cases is enough.

[Sith_Happens]

Okay, so according to that if you keep your skill rank maxed, then from 8th level and onwards you'll have a >99% success chance with any effective spell level you can attempt.

[Vhaidara]

We should import this to 3.5. It doesn't have enough free metamagic for casters to keep up with those bloody Toke of Battle weeabos, and the extra 3 ranks, and therefore 3 dice, would make this completely balanced for them to use with a fair chance of success.

[oxybe]

We should import this to 3.5. It doesn't have enough free metamagic for casters to keep up with those bloody Toke of Battle weeabos, and the extra 3 ranks, and therefore 3 dice, would make this completely balanced for them to use with a fair chance of success.

Seriously. Even just sticking to core, did you see that Fighter class and all the free feats it gets?! This, at least, starts to even the playfield.

[grarrrg]

Seriously. Even just sticking to core, did you see that Fighter class and all the free feats it gets?! This, at least, starts to even the playfield.

Yeah, but the Lightning Warrior Fighter doesn't even get a Familiar. So that helps out a little.


Not quite: at the dangerously overpowered levels, you are a bit behind: you only need 3 dice to succeed with "all 1s" for a level 1 spell.

Level 1 is quite dangerously overpowered at that...


Long drawn out math added to the 3-die-bye-bye (http://www.giantitp.com/forums/showthread.php?363930-Sacred-Geometry-and-Arithmancy&p=17883351&viewfull=1#post17883351) post.

[DarkSonic1337]

We should import this to 3.5. It doesn't have enough free metamagic for casters to keep up with those bloody Toke of Battle weeabos, and the extra 3 ranks, and therefore 3 dice, would make this completely balanced for them to use with a fair chance of success.

Besides, it's not like 3.5 even has metamagic feats that are as good as Pathfinder's to apply this to anyway, and you get less of them. So the feat would definitely be weaker for a 3.5 Sorcerer than a PF Sorcerer

[Miss Disaster]

Funny thing though ... Knowledge: Engineering is one of the least popular Knowledge skills (as it doesn't ID any creature types). So at least Sacred Geometry helps promotes a sorely underloved skill to the mass populace. C'mon people, work with me here. :smalltongue:

[Starbuck_II]

Funny thing though ... Knowledge: Engineering is one of the least popular Knowledge skills (as it doesn't ID any creature types). So at least Sacred Geometry helps promotes a sorely underloved skill to the mass populace. C'mon people, work with me here. :smalltongue:

But Red Hand of Doom uses it. Okay, only in one specific spot. It really should come up in modules/APs more often.

[OldTrees1]

Funny thing though ... Knowledge: Engineering is one of the least popular Knowledge skills (as it doesn't ID any creature types). So at least Sacred Geometry helps promotes a sorely underloved skill to the mass populace. C'mon people, work with me here. :smalltongue:

Honestly, Knowledge Dungeoneering can be substituted for a lot of Knowledge Engineering. Maybe they should be combined?

Or pull a Roy (http://www.giantitp.com/comics/oots0808.html)

[Khosan]

I've ran the numbers, and this works for all combinations of 3 dice (will post if needed, or if I get bored later).
If 3 dice do not reduce to a 0 or a 1, they WILL let you "convert" any other number into itself.
Start with all Unique 3-die combinations (56 unique), then remove any with a duplicate number, as these easily reduce to 0 (20 non-doubles). Finally, remove all the combinations that reduce to 0 or 1. This leaves 5 combinations:
1, 2, 5
1, 2, 6
3, 4, 5
3, 5, 6
4, 5, 6
We already proved 4 ,5, 6 in the example

1, 2, 5, 1 > 5-2-1-1 = 1
1, 2, 5, 2 > 5-2-2+1 = 2
1, 2, 5, 3 > 5-3+2-1 = 3
1, 2, 5, 4 > 5-4+2+1 = 4
1, 2, 5, 5 > (5*2-5)*1 = 5
1, 2, 5, 6 > (5+1)*2-6 = 6

1, 2, 6, 1 > 6/2-1-1 = 1
1, 2, 6, 2 > (6-2-2)*1 = 2
1, 2, 6, 3 > 6/3+2-1 = 3
1, 2, 6, 4 > (6-4+2)*1 = 4
1, 2, 6, 5 > (6-1)*2-5 = 5
1, 2, 6, 6 > (6*2-6)*1 = 6

3, 4, 5, 1 > 4+3-5-1 = 1
3, 4, 5, 2 > 5+3-4-2 = 2
3, 4, 5, 3 > 5*3-4*3 = 3
3, 4, 5, 4 > (5-3)*4-4 = 4
3, 4, 5, 5 > 5*4-5*3 = 5
3, 4, 5, 6 > 5+4+3-6 = 6

3, 5, 6, 1 > 5+3-6-1 = 1
3, 5, 6, 2 > 6+3-5-2 = 2
3, 5, 6, 3 > (5*3-6)/3 = 3
3, 5, 6, 4 > 6+5-4-3 = 4
3, 5, 6, 5 > 6*5/3-5 = 5
3, 5, 6, 6 > 6+6/(5-3) = 6

Apologizes if I typed a wrong symbol and missed it.


The shorter way to do this is finding combinations of 4 with no repetition (since any repetition can be brought to 0 with just those two). There's only 15 unique combinations then.

1, 2, 3, 4
1, 2, 3, 5
1, 2, 3, 6
1, 2, 4, 5
1, 2, 4, 6
1, 2, 5, 6
1, 3, 4, 5
1, 3, 4, 6
1, 3, 5, 6
1, 4, 5, 6
2, 3, 4, 5
2, 3, 4, 6
2, 3, 5, 6
2, 4, 5, 6
3, 4, 5, 6

The following sets of 3 (there are 20 unique sets of 3 with no repetition) all reduce to 0:

1, 2, 3 > 3 - 2 - 1
1, 3, 4 > 4 - 3 - 1
1, 4, 5 > 5 - 4 - 1
1, 5, 6 > 6 - 5 - 1
2, 3, 5 > 5 - 3 - 2
2, 3, 6 > (6 / 2) - 3
2, 4, 6 > 6 - 4 - 2

And at least one of these is contained within all but one of the sets of 4 listed above. The only one that doesn't is 3, 4, 5, 6, which still reduces to 0 (6 + 3 - 4 - 5), you just need all 4 numbers to do it.

So for any given set of 4, there always exists a solution that brings it to 0. There are 126 possible sets of 4 (counting repetition this time), of those 111 contain a pair (or triplet or quadruplet) and can be solved with 2, of the remaining 15, 14 can be solved with 3 and the remaining 1 can be solved with all 4.

EDIT: Oh, and for sets of 3, there are 56 unique rolls. 36 of those contain a pair and can be reduced. 7 of the remaining 20 can be reduced to 0 using all 3. The last 13 can't be brought down to 0 without the addition of a 4th roll.

So if you roll 3 dice there's a ~75% chance you'll be able to equate them to 0.

[grarrrg]

MORE NUMBERS!

I ran the numbers for 1st level success with various numbers of dice.
Note: I will try to list results that DON'T work, on the chance I missed something. If there are very few that do work, I will list those instead.
Also note: It's hard/weird to write out strings of numbers, as I'm really not sure how to separate them, hopefully it's understandable.

1st level spells (Primes: 3, 5, 7)
1d6 has a 1/3 (2/6) chance to succeed. You must roll a 3 or a 5
2d6 has a 5/9 (20/36) chance to succeed. All "doubles" fail (1ea.) as do 24, 35, 45, 46, 56 (2ea.)
3d6 has a 11/12 (198/216) chance to succeed. Failures are 155, 166 (3ea.) and 145, 156 (6ea.)
4d6 always succeeds

2nd level spells (Primes: 11, 13, 17)
2d6 has 1/18 (2/36) chance to succeed. You must roll a 5,6 (2ea.)
3d6 has 19/36 (114/216) chance to succeed. Fairly even split, 24 unique combo's work, 32 unique combo's fail.
Anything with a Pair of 1's or a Pair of 2's
All Triplets
Combo's where all numbers are Even
123, 124, 133, 135, 145, 255, 336, 366, 455, 456, 556
4d6 has 1247/1296 chance to succeed. Failures: 1111, 1112, 1113, 1114, 1122, 1123, 1222, 1555, 2222, 2224, 2444, 3333
5d6 has 7753/7776 chance to succeed. Failures: 11111, 22222, 55555 (1ea.), 11112, 11113 (5ea.), 11122 (10)
6d6 has 46649/46656 chance to succeed. Failures: 111111 (1), 111112 (6)
7d6 suspect the only failure will be all 1's.

3rd level spells (Primes: 19, 23, 29)
4d6
5d6 has 1921/1944 (7684/7776) chance. Fails with all combinations of 1's and/or 2's (32 total), 11113, 11114, 11115, 22224 (5ea.), 11123, 11124 (20ea.)
6d6

4th level spells (Primes:)

5th level spells (Primes:)

6th level spells (Primes:)
4d6 85/648 (170/1296) meschlum post.

7th level spells (Primes:)
4d6 41/648 (82/1296) meschlum post.

8th level spells (Primes:)
4d6 has 5/108 (60/1296) meschlum post.

9th level spells (Primes: 101, 103, 107)
4d6 has 5/108 (60/1296) chance. Succeeds only with 4456, 1455, 3455, 1366, 3566 (12 ea.)


For completing the table (if we get that far), I'd recommend starting with the bare minimum number of dice (the "Hail Mary (http://www.giantitp.com/forums/showthread.php?363930-Sacred-Geometry-and-Arithmancy&p=17869734&viewfull=1#post17869734)"s from last page, and the lowest level you could cast the spell (i.e. 5th spells would be 9 ranks).
Then fill in where the bigger gaps are. If, as an example, 10 dice with 9th level spells is a 99% success, then it doesn't really pay to check 11, 12, or 13 dice, as they're all going to be fractions of a percent different.

Spaces with a "?" are recommended start points, although any/all data will be accepted.
Spaces to the left of 0% are also 0%, spaces to the right of 100% are also 100%.



1d
2d
3d
4d
5d
6d
7d
8d
9d
10d
11d
12d
13d
14d


1st
33.33
55.55
91.67
100%












2nd
0%
5.56
52.78
96.22
99.70
99.98

100%








3rd

0%
16.67

98.82



100%







4th

0%
5.56



?



100%





5th


0%
?




?


100%




6th


0%
13.12


?



?

100%



7th


0%
6.33



?




100%



8th


0%
4.63




?




100%


9th


0%
4.63




?




100%




The shorter way to do this is finding combinations of 4 with no repetition (since any repetition can be brought to 0 with just those two)....
EDIT: Oh, and for sets of 3, there are 56 unique rolls. 36 of those contain a pair and can be reduced. 7 of the remaining 20 can be reduced to 0 using all 3. The last 13 can't be brought down to 0 without the addition of a 4th roll.

So if you roll 3 dice there's a ~75% chance you'll be able to equate them to 0.

I think you missed part of the point here. We are not just looking for a 0.

If we can get a solution and have some number of leftover dice, can we automatically "ignore" those extra dice, or not?
There are 3 ways:
Make the extras equal 0, then we can just "+0" our solution and we're good.
Make the extras equal 1, then we can just "*1" our solution and we're good.
Make the extras "condense" into an existing number.

It was already proven that 4 was the minimum to get a 0 or 1 and just disappear outright. I went through and showed that 3 dice either disappear outright, or condense into an existing number.

[Khosan]

I think you missed part of the point here. We are not just looking for a 0.

If we can get a solution and have some number of leftover dice, can we automatically "ignore" those extra dice, or not?
There are 3 ways:
Make the extras equal 0, then we can just "+0" our solution and we're good.
Make the extras equal 1, then we can just "*1" our solution and we're good.
Make the extras "condense" into an existing number.

It was already proven that 4 was the minimum to get a 0 or 1 and just disappear outright. I went through and showed that 3 dice either disappear outright, or condense into an existing number.

I just prefer reducing to 0. Feels cleaner to me, less mathy. It's almost purely addition and subtraction (there's one division in there).

I did leave out 1s in that last bit, which would be a bit misleading. Adding in reductions to 1, the chance of rolling 3 die and being able to reduce them to 0 or 1 is...51/56. Just about 90%. The remaining 5 combinations would require one extra die, like you mentioned. And given that this is just what's left over, there's at least a small chance you could get the same result by just rearranging the actual answer so you end up with a different set of 3.

[grarrrg]

I just prefer reducing to 0. Feels cleaner to me, less mathy.

Were all about the "more mathy" here.
We're trying to find the difference between "probably works" and "definitely works", thus MATH is involved, and simple shortcuts are frowned upon, unless they have been tested and can be relied upon.


...The remaining 5 combinations would require one extra die, like you mentioned. And given that this is just what's left over, there's at least a small chance you could get the same result by just rearranging the actual answer so you end up with a different set of 3.

"Probably" getting rid of 3 extra dice is nice and all, but "Knowing" you can get rid of 3 extra dice is much better.
And we have proven you KNOW you can get rid of 3 extra dice.


the chance of rolling 3 die and being able to reduce them to 0 or 1 is...51/56. Just about 90%.

[MATH!]
Technically 31/36 (or more accurately 186/216) for 86.11% chance to reduce to 0 or 1.
As some unique combinations are truly unique (1/216 for the triples), with 2 groups being more common (3/216 for the "pairs" and 6/216 for the "all different")
[/MATH!]

[meschlum]

So let's look at three dice, then. Mostly because it's easier.



Level 3 requires 19, 23, or 29, none of which are possible via sums of 3 dice.

If a division is used, there is at most one 'free' die left, with a maximum value of 6. Since we're looking for primes, the divided term needs to be added to the remaining die (multiplying won't help), so this gives a maximum of 12.

Therefore, there is a multiplication and an addition (or a subtraction) (two multiplications won't give a prime), and the addition (or subtraction) happens after the multiplication.

19 + 6 = 25, which can be reached via 5 * 5 only, so 5, 5, 6 is a solution
19 + 5 = 24, which can be reached via 4 * 6 only, so 4, 5, 6 is a solution
19 + 4 = 23, which is a prime and cannot be reached
19 + 3 = 22, which cannot be reached with two dice
19 + 2 = 21, which cannot be reached with two dice
19 + 1 = 20, which can be reached via 4 * 5 only, so 1, 4, 5 is a solution
19 - 1 = 18, which can be reached via 3 * 6 only, so 1, 3, 6 is a solution
19 - 2 = 17, which is a prime and cannot be reached
19 - 3 = 16, which can be reached via 4 * 4 only, so 3, 4, 4 is a solution
19 - 4 = 15, which can be reached via 3 * 5 only, so 3, 4, 5 is a solution
19 - 5 = 14, which cannot be reached with two d6.
19 - 6 = 13 which is a prime and cannot be reached

Since solutions for one prime are solutions for all, the range covered when starting at 29 is 23 to 35, so it covers all options for 23 as well.

26: not possible with two dice
27: not possible with two dice
28: not possible with two dice
29: prime
30: 5 * 6 - 1 is a solution
31: prime
32: not possible with two dice
33: not possible with two dice
34: not possible with two dice
35: not possible with two dice

So the solutions for level 3 are:

1, 3, 6 (6 values)
1, 4, 5 (6 values)
1, 5, 6 (6 values)
3, 4, 4 (3 values)
3, 4, 5 (6 values)
4, 5, 6 (6 values)
5, 5, 6 (3 values)

So there are 36 outcomes where level 3 is possible, so a probability of 1/6 (16.7%)

Level 4 uses 31, 37, 41, so the range of numbers covered is 25 to 47.

There are two solutions for values in the 25 to 35 range (5, 5, 6 and 1, 5, 6).

The maximum with two dice is 36, so this adds a single extra solution (1, 6, 6).

The rolls that succeed for level 4 are therefore:

1, 5, 6 (6 values)
1, 6, 6 (3 values)
5, 5, 6 (3 values)

So Level 4 succeeds with 3 dice with probability 12 / 216 = 1 / 18 (5.56%)



We get, with three dice:

Level 3:1/6 (16.7%)
Level 4: 1/18 (5.56%)


And the reasoning used to solve quickly with three dice breaks down when more are used, since division can be more important. On the other hand, my solutions don't use much division besides getting 1 from pairs, so it might work.

[grarrrg]

The 2's are effectively done. 5 dice is a 99.7% chance, and 6 dice is 99.98%
I don't think there is a need to figure out 7 dice.


We get, with three dice:

Level 3:1/6 (16.7%)
Level 4: 1/18 (5.56%)

And the reasoning used to solve quickly with three dice breaks down when more are used, since division can be more important.

I like the method behind this, but it quickly becomes worthless, as you stated.
Applied to 4 dice at 9th level our target values are:
4*4*6 = 96
4*5*5 = 100
3*6*6 = 108

Double 4 just barely gets you to 101 with a 5
Double 5 gets you 101 with a 1 and 103 with a 3
Double 6 gets 107 with a -1 and 103 with a -5
Each of these combinations appears 12 times each for 5/108 (60/1296)

[Gemini476]

So if you were to actually use Sacred Geometry in a game, obviously staying at enough ranks of K(Eng) to auto-succeed and not have to slow down the game with math, what metamagic feats should you take?

Let's have a hypothetical human metawizard who really likes metamagic. His feats are Scribe Scroll, six [Metamagic] feats, and nine Sacred Geometry feats.
That's a total of 24 metamagics, 18 of which he does not need to learn the prerequisites for before he takes them.

So which should he choose? Heighten Spell, obviously, but what else?

And before you say "all of them", I checked and there are actually 39 [Metamagic] feats so you'll miss out on 15 of them.

[Snowbluff]

I said so earlier. Quicken and Echoing. Quicken spell means you won't lose an important action if you screw up, and it's generally hard to use without making it free.

Echoing buffs the crap out of your longevity as a caster. Double all of my spells level 6 and below? Yes please.

[Erik Vale]

Still/Silent [Duh], Quicken [Speed], Highten [Because free DC booster], Echooing [Boost Longgevity]

[DarkSonic1337]

Heighten

Persistent Spell to make everyone save twice against your stuff, Dazing Spell to add Dazes to stuff, Echoing Spell for longevity, Extend Spell all of your buffs, Lingering Spell for the lulz, Quicken Spell for the obvious action economy breakage, Reach Spell because some touch spells would be really good without the "touch" part, Maximize Spell for your Calcific Touch/Enervation/Whatever (empower too), Selective Spell for your party friendly battlefield control, Silent and Still spell for panic buttons, Toppling Spell for more battlefield control, Umbral Spell+Shadow Grasp for fun darkness stuff and every spell you cast entangles.

Maybe add on Widen and Enlarge too.

[Kantolin]

Do note that quicken, in this case, only allows you to use the spell in its normal activation time rather than as a swift action or the normal full-round action. Which would thus be fairly critical for a spontaneous caster, but not so helpful for a prepared caster; for a wizard I'd probably either skip it or take it as a normal feat.

(...unless I'm misreading Sacred Geometry, or unless you have the metamagic down to 0 in which may as well)

[Snowbluff]

Do note that quicken, in this case, only allows you to use the spell in its normal activation time rather than as a swift action or the normal full-round action. Which would thus be fairly critical for a spontaneous caster, but not so helpful for a prepared caster; for a wizard I'd probably either skip it or take it as a normal feat.

(...unless I'm misreading Sacred Geometry, or unless you have the metamagic down to 0 in which may as well)
*shakes head*
1) You negate the increased casting time, as per Sacred Geometry.
2) You apply Quicken Spell, which works with full round action spells anyway, making it a swift action.

Quicken works with Spontaneous Casters in the feat text. (http://www.d20pfsrd.com/feats/metamagic-feats/quicken-spell-metamagic---final)

[Kantolin]

Quicken works with Spontaneous Casters in the feat text.

Wow, you learn something new every day. :smallsmile: Definitely a positive change.



1) You negate the increased casting time, as per Sacred Geometry.
2) You apply Quicken Spell, which works with full round action spells anyway, making it a swift action.

Are... you sure this works? It seems to imply 'applying quicken spell negates the increased time'. But then again, I could see it interpreted, 'And then works normally', it just doesn't seem like what it's saying.

I mean Sacred Geometry is pretty whacked out with or without quicken, so I can't say what RAI is from a balance standpoint as there isn't any of this 'balance' stuff. :smallsmile: And it's perfectly reasonable that my knee-jerk reaction is to aim weaker.

Either way, if it does work then totally quicken. If it doesn't work, then spontaneous people will want to quicken so it's not two full round actions, while prepareds can probably go without it.

[Snowbluff]

*ahem* If a human with the Human Heritage Feat become a Necropolitan, what happens to it when you try to rebuke it?

Nothing! It's not Human, but the feat says it still counts as such. The same applies here. Sure, it may say one thing, but that does not mean that the other feat or ability still doesn't work as normal.

[Raven777]

I'm pretty sure that the intent of the feat is for the player to calculate the prime combinations from his roll in real time and show it to the table, not just point to a proof.

[Snowbluff]

I'm pretty sure that the intent of the feat is for the player to calculate the prime combinations from his roll in real time and show it to the table, not just point to a proof.

Oh, it is.

1) Intent is useless.
2) We know it works.
3) Since we know it works, why not just skip the slow, tiresome task of running the numbers in real time?

[Raven777]

I don't know... I mean... I feel... I love me some Paragon Surge, but this? This just feels... wrong.

[Snowbluff]

I don't know... I mean... I feel... I love me some Paragon Surge, but this? This just feels... wrong.

But it's so right, Raven. Join me, and as Paladin and Pony we shall rule the realm! Let me show you the true power of the dork side! :smalltongue:

[grarrrg]

I'm pretty sure that the intent of the feat is for the player to calculate the prime combinations from his roll in real time and show it to the table, not just point to a proof.

The main issue with this is that the "genius" designer at Paizo apparently couldn't be bothered to run the actual numbers, and thus we wind up with a Feat that is a huge power-boost for the classes that are already the strongest.

In a game with a level cap of, oh, 10, the feat is "ok". Meta-boosts up to 3rd level are 100%, but there's still a chance to fail on your highest level slots (not much of a chance...but a chance).
In a game that goes to level 13 or higher, it's just pure stupid-overpowered-ness, as you can not fail with 14 ranks.

The real 'intent' was "Hey guys, I have a stupid idea for a feat that requires lots of numbers and stuff! The players will probably look at it and go 'I'm not doing math', so let's throw in a HUGE bonus if they actually do!"


Also, why'd they skip 2? 2 is a perfectly good Prime Number.

[Snowbluff]

The real 'intent' was "Hey guys, I have a stupid idea for a feat that requires lots of numbers and stuff! The players will probably look at it and go 'I'm not doing math', so let's throw in a HUGE bonus if they actually do!"
Players not doing math... I can't even... grarrrg help me! D:


Also, why'd they skip 2? 2 is a perfectly good Prime Number.

Maybe the designer doesn't know what a prime number is?

[Kalaska'Agathas]

Maybe the designer doesn't know what a prime number is?

But then how did they get the rest of them? I mean, if they're picking from a list of primes, it's not like two isn't going to be on there.

[Vhaidara]

But then how did they get the rest of them? I mean, if they're picking from a list of primes, it's not like two isn't going to be on there.

Lucky shots on a dart board?

[Snowbluff]

But then how did they get the rest of them? I mean, if they're picking from a list of primes, it's not like two isn't going to be on there.

Lucky guesses.

EDIT: Wizarded.

[meschlum]

Also, why'd they skip 2? 2 is a perfectly good Prime Number.

Because that way they can announce an extended version of the feat which allows Level 0 spells, using 0, 1, and 2! Precipitating fights over whether 1 is prime or not, and adding to the chain of actively disadvantageous feats that build on the initial awesome one.

The other explanation is that managing to upgrade a cantrip to a level 1 spell was judged to be so powerful, it couldn't be allowed to happen 50% of the time at level 1 (when you can't get the feat anyway, since it requires 2 ranks (and so 2 dice). Or maybe they went so far as to check the results with three dice, and discovered that you can get 2, 3, 5 all the time with three dice?



We'll start high and see how low we can go.

The reasoning for three dice can be extended somewhat: with two dice, the maximum you can get is 36, so any level of spells that requires more than that has to use at least three dice to get close. With three dice, you can go to 216, which is far more than you can use, or (6 + 6) * 6, indicating that any number over 72 must be rolled using the product of three dice, with the last one added or subtracted.

That means level 8 and level 9 spells should be easier to solve.

Level 9: possible three die products are in the range from 95 to 113, and we get:

95: 19 * 5, not possible with a three die product
96: 3 * 2^5. If there is a 3, then we need 32 with two dice, which is not possible - therefore, there is a 6, and we need 16 with two dice. The only solution is 4, 4, 6 (+5)
97: prime
98: 2 * 7^2, not possible with three dice
99: 3^2 * 11, not possible with three dice
100: 2^2 * 5^2. If there is a 2, we need 50 with two dice, which is not possible - therefore, there is a 4, and we need 25 with two dice. The only solution is 4, 5, 5 (+1 or +3)
101: prime
102: 2 * 3 * 17, not possible with three dice
103: prime
104: 2^3 * 13, not possible with three dice
105: 3 * 5 * 7, not possible with three dice
106: 2 * 53, not possible with three dice
107: prime
108: 2^2 * 3^3. If there is a 2, we need 54 with two dice, which is not possible. If there is a 4, we need 27 with two dice, which is not possible. Therefore, there is a 6, and we need 18 with two dice. The only solution is 3, 6, 6 (-1 or -5)
109: prime
110: 2 * 5 * 11, not possible with three dice
111: 3 * 37, not possible with three dice
112: 2^4 * 7, not possible with three dice
113: prime

So for level 9, the only four die solutions are:

1, 3, 6, 6
1, 4, 5, 5
3, 4, 5, 5
3, 5, 6, 6
4, 4, 5, 6

For a total of 60 / 1296, or 5 / 108 ~4.63%

Level 8 has us range from 77 to 103, and we get:

77: not possible with three dice
78: not possible
79: prime
80: 2^4 * 5, so there must be a 5, and 4 * 4 * 5 is the only solution (+3)
81: 3^4, not possible with three dice
82: not possible
83: prime
84: hasa 7 factor, not possible
85: not possible
86: not possible
87: not possible
88: not possible
89: prime
90: 2 * 3^2 * 5, so there must be a 5, and 3 * 5 * 6 is the only solution (-1)
91: not possible
92: not possible
93: not possible
94: not possible

96 is possible, and uses 1 only
100 is possible and uses 3 only

So the possible solutions are

1, 3, 5, 6
1, 4, 4, 6
3, 4, 4, 5
3, 4, 5, 5

For a total of 60 / 1296 as well, so 4.63%


Level 7 can be reached via three die products, and via (6 + 6) * 6 = 72, or (6 + 5) * 6 = 66. Taking a lower multiplier limits the result to 60, which is out of range for the remaining die. The sum of two products is 72 at most, but to get a prime we can use 25 + 36 = 61 at best, which does not work with Level 7.

Three die products range from 65 to 85, so the following work:

80 (4, 4, 5 only, -1)
75 (3, 5, 5 only, -2, +/-4)
72 (2, 6, 6 and 3, 4, 6, +/- 1)

Solutions with 72 are +/-1, and with 66 are +5

Therefore, the following are the Level 7 solutions with four dice:

1, 2, 6, 6
1, 3, 4, 6
1, 4, 4, 5
1, 6, 6, 6
2, 3, 5, 5
3, 4, 5, 5
5, 5, 6, 6

For a total of 82 / 1296 ~6.327%


Level 6 has more options: three die product, (4 + 5) * 6, (3 + 6) * 6, (4 + 6) * 6, (5 + 5) * 6, (5 + 6) * 6, (6 + 6) * 6, (6 + 6) * 5, (6 + 5) * 5 as well as 5*5 + 6*6

Three die products must range from 53 to 73, so the options are:

54 (3, 3, 6 only, +5)
60 (2, 5, 6 and 3, 4, 5, +/-1)
64 (4, 4, 4 only, +/-3)
72 (2, 6, 6 and 3, 4, 6, -5)

Many of the products above can also be generated with the sum then product approach, as well as 55 (+4 and +6) and 66 (+1 and -5)

3, 6, 6 (+5)
4, 5, 6 (+5)
4, 6, 6 (+/-1)
5, 5, 6 (+/-1 with 60, +4 and +6 with 55)
5, 6, 6 (+/-1 with 60, +1 and -5 with 66)
6, 6, 6 (-5)

The following are level 6 solutions with four dice:

1, 2, 5, 6
1, 3, 4, 5
1, 4, 6, 6
1, 5, 5, 6
1, 5, 6, 6
2, 5, 6, 6
3, 3, 5, 6
3, 4, 4, 4
3, 4, 5, 6
3, 5, 6, 6
4, 5, 5, 6
5, 5, 6, 6
5, 6, 6, 6

For a total of 170 / 1296 ~13.1% chance of success with Level 6.



Four dice:

Level 6: 170 / 1296 ~ 13.1%
Level 7: 82 / 1296 ~ 6.33%
Level 8: 60 / 1296 ~4.63%
Level 9: 60 / 1296 ~4.63%

So with four dice, your odds of success are comparable to (or better than) the chance of scoring a crit in the worst case scenario (5% of rolling a 20, then need to confirm).

[grarrrg]

Players not doing math... I can't even... grarrrg help me! D:

ROLE players, not ROLL players :smallwink::smallbiggrin:
*hands snowbluff a paperbag to breathe into*


Four dice:

Level 6: 170 / 1296 ~ 13.1%
Level 7: 82 / 1296 ~ 6.33%
Level 8: 60 / 1296 ~4.63%
Level 9: 60 / 1296 ~4.63%

So with four dice, your odds of success are comparable to (or better than) the chance of scoring a crit in the worst case scenario (5% of rolling a 20, then need to confirm).

I ran the numbers for 5 dice with 3rd level.
98.82% chance of success.
7684/7776 chance. Fails with all combinations of 1's and/or 2's (32 total), 11113, 11114, 11115, 22224 (5ea.), 11123, 11124 (20ea.)

While we only have limited data, it's currently looking like the numbers rise really fast for the first few dice, and then s..l..o..w..l..y ease into 100%

[Chronos]

There's yet another reason why this feat is really bad game design. Not only does it mix real-world ability up with game ability, and not only does it horribly bog down play, and not only does it involve an activity completely unlike anything the players signed up for playing, and not only does it not match its fluff, being neither sacred nor geometric, and not only is it horribly overpowered... But it's not immediately obvious just how horribly overpowered it is, making it more likely that a DM would unwittingly allow it for his games.

Sometimes when I see these horribly overpowered combinations, or confusingly-worded abilities, I half suspect that the authors who created them were just trying to foist them off on their own DMs for purposes of their own personal powergaming. See, it's published! It's a real rule! I totally get to use it in game!

[Elderand]

and not only does it involve an activity completely unlike anything the players signed up for playing

While I agree with the rest of your post, this I object to.

It's a feat, the player chose to take it; therefore they, in fact, did sign up for exactly this when they take it.

[georgie_leech]

While I agree with the rest of your post, this I object to.

It's a feat, the player chose to take it; therefore they, in fact, did sign up for exactly this when they take it.

It is fundamentally different from pretty much every other use of the dice in the game though. New subsystems are one thing, but this would be closer to a house rule that your skill with Darts was determined by actually throwing darts at a dartboard.

[Elderand]

It is fundamentally different from pretty much every other use of the dice in the game though. New subsystems are one thing, but this would be closer to a house rule that your skill with Darts was determined by actually throwing darts at a dartboard.

I'm not arguing it's not different, I'm arguing that you can't say the player didn't sign up for it because they have to chose to take that feat.

[meschlum]

Why not have a look at six dice? It's enough to have a chance of success for more or less everything, low enough that you're only guaranteed to work with Level 1 spells, and few enough dice that plotting out combinations isn't completely insane.



Starting with level 9, because (presumably) the odds of success will be higher for lower levels.

All 1s fail, since fourteen are needed.

With only 1s and 2s, there are no solutions, since the highest value would be 64 (2 * 2 * 2 * 2 * 2 * 2)

With only 1s, 2s, and 3s, some solutions are possible:

3 * 3 * 3 * 2 * 2 - 1
3 * 3 * 3 * (3 + 1) - 1

These are the only solutions using products to get 108, since replacing any of the 3s in the product would require an extra die at least (3 becomes 2 + 1), and both options for getting 4 (two 2s and a 1 and a 3) are used.

To get 100 with 1s, 2s, and 3s, we need a 4 (two dice) and two 5s (two dice each), so six or more dice, which is not allowed.

The only six die solutions using 1s, 2s, and 3s (729 combinations) are:

1, 1, 3, 3, 3, 3 (15 combinations)
1, 2, 2, 3, 3, 3 (60 combinations)

Therefore, there are 654 combinations using 1s, 2s, and 3s only that are not solutions for level 9 with six dice.

If the dice are 1s to 4s, more solutions turn up.

With 4, 3, 3, 3, the remaining two dice just need to combine to get 1, 5, or 7 and there is a solution, so pairs always work.

1, 2: works
1, 3: may not work
1, 4: works
2, 3: works
2, 4: may not work
3, 4: works

We need at least five dice to get 100 (4 * (2 + 3) * (2 + 3)), so the uncertain combinations do not allow 100 to be generated as two 2s are needed to generate 100, and neither uncertain combination includes two 2s. Therefore, the following are NOT solutions:

1, 3, 3, 3, 3, 4 (30 combinations)
2, 3, 3, 3, 4, 4 (60 combinations)

This means that the following are all solutions:

1, 1, 3, 3, 3, 4 (60 combinations)
1, 2, 3, 3, 3, 4 (120 combinations)
1, 3, 3, 3, 4, 4 (60 combinations)
2, 2, 3, 3, 3, 4 (60 combinations)
2, 3, 3, 3, 3, 4 (30 combinations)
3, 3, 3, 3, 3, 4 (6 combinations)
3, 3, 3, 3, 4, 4 (15 combinations)

With 4, 3, 3 and trying to generate 108, the remaining three dice need to provide a 3 and a 1, 5, or 7. Since 4, 3, 3, 3 has been covered, this means that the 3 is generated with 1 + 2 or 4 - 1, and the remaining die is a 1.

So with 4, 3, 3, the following are solutions:

1, 1, 2, 3, 3, 4 (180 combinations)
1, 1, 3, 3, 4, 4 (90 combinations)

With 4, 3, 3 and trying to generate 100, the remaining three dice must be 1, 2, 2, or 3, 2, 2. Since we have already covered cases with three 3s, the latter is not counted, and the following is a solution as well.

1, 2, 2, 3, 3, 4 (180 combinations)

With 4, 3 exactly, we need to use four more dice to generate the two missing 3s, so there are no solutions going to 108. We need five dice in all to get 100 with 4 * (3 + 2) * (1 + 4), and the remaining die can be a 1 or a 3. If the remaining die is a 3, then 4 * (1 + 4) * (1 + 4) is a solution for 100. The following are therefore solutions using 4s:

1, 1, 2, 3, 4, 4 (180 combinations)
1, 1, 3, 4, 4, 4 (60 combinations)
1, 2, 3, 3, 4, 4 (180 combinations)

If there are 4s and no 3s, it is not possible to reach 108 (two dice are needed for a 3, so six extra in all). To get 100, we need to use 4 * (4 + 1) * (4 + 1), and the last die can be a 1 or a 3 - and so must be a 1.

1, 1, 1, 4, 4, 4 (20 combinations)

There are therefore a total of 1,241 combinations that solve Level 9 with dice in the 1s to 4s (out of 4096 total), and 2855 that are not solutions.


If there is at least one 5, the number of solutions increases again.

With four or more 5s, we have (5 + 5) * (5 + 5), and the remaining two dice must add up to 1, 3 or 7. This always works if there is a pair or if the dice are 1 or 3 apart, leaving the following:

1, 3: works (1 * 3)
1, 5: may not work
2, 4: may not work
2, 5: works (2 + 5)
3, 5: may not work

In cases where there are five 5s, there is a 1 or a 3, and 5 * 5 * 5 - 5 * 5 = 100, so there is a solution.

The remaining case is 2, 4, 5, 5, 5, 5, which can be solved with 5 * 5 * 4 + 2 + 5/5.

So all combinations with four or more 5s are solutions.

If there are exactly three 5s, we can get 100 with 5 * 5 * (5 - 1), so there is a solution in all cases where there is a 1 and the other two dice add up to 1, 3, or 7. This means that the only option in this case is 1, 2, 4, 5, 5, 5, which is solved with 5 * 5 * 4 + 5 + 2 * 1. So if there are exactly three 5s and one or more 1s, there is always a solution.

If there are exactly three 5s and one or more 4s, we have 5 * 5 * 4 = 100, so the remaining three dice (including exactly one 5) must add up to 1, 3, or 7.

1, 1, 5: works
1, 2, 5: works
1, 3, 5: works
1, 4, 5: works
2, 2, 5: works
2, 3, 5: works with (2+3)/5
2, 4, 5: works
3, 3, 5: works
3, 4, 5: may not work: 3, 4, 4, 5, 5, 5 is solved with 5 * 5 * 5 - 3 * (4 + 4), so it works
4, 4, 5: works

Therefore, if there are three 5s, there is a solution if any of the other dice are 1s or 4s.

2, 2, 2, 5, 5, 5: works (5 * 5 * 2 * 2 + 5 + 2)
2, 2, 3, 5, 5, 5: works (5 * 2 + 2) * (5 + 3) + 5 = 101
2, 3, 3, 5, 5, 5: works (5 * 2 + 3) * (5 + 3) + 3 =107
3, 3, 3, 5, 5, 5: works (5 + 5) * (5 + 3 + 3) - 3 = 107

Therefore, there is always a solution if there are three or more 5s.

With exactly two 5s and a 4, we need the remaining three dice to be equal to 1, 3, or 7. If one of the dice is a 1, there is always a solution unless the other dice are a 2 and 4, which is a solution since 1 + 2 + 4 = 7.

Therefore, if there are exactly two 5s and at least one 4, there is always a solution if any of the remaining dice is a 1.

If any of the remaining dice are a pair, then we have 1+2, 1*3, or 4-1, so there is always a solution. Therefore, the three remaining dice must be 2, 3, 4, and there is a solution since 2 + 4 - 3 = 3.

So there is always a solution with two 5s if there are any 4s.

With two 5s and any two dice that add up to 4, there is a solution if the remaining two dice add up to 1, 3, or 7 - which is always the case since they are in the 1 to 3 range.

Two dice in the 1 to 3 range do not add up to 4 only if they are 1, 1 or 1, 2 or 2, 3 or 3, 3. Since there are four dice in all that are not 5s, we have only the following cases with two 5s that may not be solutions:

1, 1, 1, 1, 5, 5: not a solution (15 cases)
1, 1, 1, 2, 5, 5: 5 * 5 * (2 + 1 + 1) + 1 is a solution
2, 3, 3, 3, 5, 5: 5 * 5 * (3 + 3 - 2) + 3 is a solution
3, 3, 3, 3, 5, 5: (3 + (3 * 3)) * (5 + 3) + 5 is a solution

So with two or more 5s, there are only 15 combinations that are not solutions.

If there is exactly one 5, there are yet more cases to check.

If there is a 5 and one or more 4s, then there is a solution when there are two dice that add up to 5 and two other dice that add up to 1, 3, or 7. This means that the second set of dice must be 2, 4 if there is no solution.

If the first set of two dice (adding up to 5) is 1, 4, then we need to solve 1, 2, 4, 4, 4, 5: (5 + 4 + 4) * 2 * 4 - 1 is a solution.

If the first set of dice is 2, 3, then we need to solve 2, 2, 3, 4, 4, 5: (5 + 4) * 3 * 4 - 2/2 is a solution.

So there is always a solution if any two dice add up to 5. Therefore, if there are any 1s, there are no 4s, and 2s exclude 3s by the same reasoning.

If three dice add up to 5 (1, 1, 3 or 1, 2, 2), then the last die must be a 2 or a 4. Since there is a 1, the last die cannot be a 4, so we need to check the following.

1, 1, 2, 3, 4, 5: solved by (5 + 1) * (4 + 2) * 3 - 1
1, 2, 2, 2, 4, 5: solved by (4 * 2 + 2) * 2 * 5 + 1
The remaining dice combinations are:

1, 1, 1, 1, 4, 5: no solutions (30 combinations)
1, 1, 1, 2, 4, 5: no solutions (120 combinations)
2, 2, 2, 2, 4, 5: no solutions (30 combinations)
2, 2, 2, 4, 4, 5: solved by (4 + 2) * 4 * 2 * 2 + 5
2, 2, 4, 4, 4, 5: solved by (4 + 2) * 4 * 4 + 5 + 2
2, 4, 4, 4, 4, 5: solved by (4 + 4 + 4) * 4 * 2 + 5
3, 3, 3, 3, 4, 5: solved by (5 + 3) * (3 * 3 + 4) - 3
3, 3, 3, 4, 4, 5: solved by 5 * 4 * (4 + 3 / 3) + 3
3, 3, 4, 4, 4, 5: solved by 3 * 3 * (4 + 4 + 4) - 5
3, 4, 4, 4, 4, 5: solved by 5 * 4 * (4 + 4 / 4) + 3
4, 4, 4, 4, 4, 5: solved by (4 + 4) * (4 + 4 + 4) + 5

So if there is one 5 and one or more 4s, there are 180 combinations that are not solutions.

If there is a 5 and three or more 3s, then the following combinations are present:

1, 1, 3, 3, 3, 5: solved by 3 * 3 * 3 * (5 - 1) - 1
1, 2, 3, 3, 3, 5: solved by 5 * (2 + 3) * (3 + 1) + 3
1, 3, 3, 3, 3, 5: solved by (5 * 3 - 3) * 3 * 3 - 1
2, 2, 3, 3, 3, 5: solved by 2 * 2 * 3 * 3 * 3 - 5
2, 3, 3, 3, 3, 5: solved by (5 + 3 + 3) * 3 * 3 + 2
3, 3, 3, 3, 3, 5: solved by (3 * 3 + 3) * 3 * 3 - 5

So there is always a solution with one 5 and three or more 3s.

With one 5 and two 3s, there are three dice that are 1s and 2s.

1, 1, 1, 3, 3, 5: no solution (60 combinations)
1, 1, 2, 3, 3, 5: solved by (5 + 1) * 3 * 3 * 2 - 1
1, 2, 2, 3, 3, 5: solved by (5 * 2 - 1) * 3 * 3 + 2
2, 2, 2, 3, 3, 5: solved by 5 * (2 + 3) * (2 + 2) + 3

So there are 60 combinations that are not solutions with one 5 and two 3s.

With one 5 and one 3, there are four dice that are 1s and 2s.

1, 1, 1, 1, 3, 5: no solution (30 combinations)
1, 1, 1, 2, 3, 5: no solution (120 combinations)
1, 1, 2, 2, 3, 5: no solution (180 combinations)
1, 2, 2, 2, 3, 5: solved by 5 * (3 + 2) * 2 * 2 + 1
2, 2, 2, 2, 3, 5: no solution (30 combinations)

With a 5 and no 3s or 4s, the other dice are 1s and 2s. These have a maximum value of 32 (2 * 2 * 2 * 2 * 2), or 16 if four dice are used. Since 5 * 16 = 80, there is no way to get a total close to the target values for level 9 in this case. Since there are five dice, this gives 192 combinations in all that are not solutions (32 combinations of 1s and 2s, times six positions for the 5)

Therefore, when all dice are 5 or less and there is one or more 5s, there are 807 combinations that are not solutions. This gives a total of 3662 combinations that are not solutions, out of 15625.


And on to 6s!

With four or more 6s, we can get 6 * (6 + 6 + 6), so we just need a 1, 5, or 7 with the remaining two dice. The possible combinations for the two remaining dice are therefore:

1, 3: solved by 6 * 6 * 3 - 1 * 6 / 6
2, 4: solved by (6 * 6 * 6 - 6) / 2 - 4
2, 6: solved by 6 * (6 * 2 + 6) - 6 / 6
3, 5: solved by 6 * 3 * 6 - 5 * 6 / 6
3, 6: solved by 6 * 3 * 6 - 6 - 6 / 6
4, 6: solved by 6 * (6 * 4 - 6) - 6 / 6

So there is always a solution if four or more 6s are rolled.

With three 6s and a 2, we have 6 * 6 * 6 / 2, so we again need a 1, 5, or 7 - and none of the remaining two dice can be a 6, so the options are:

1, 3: solved by 6 * 6 * 3 + 2 - 6 - 1
2, 4: solved by (6 * 6 * 6 - 2) / 2 - 4
3, 5: solved by 6 * 6 * 3 -6 + 2 + 3

So if there are three 6s, there cannot be any 2s.

With three 6s and a 3, we have 6 * 6 * 3 - 6 (102), so we need a 1 or a 5 with two dice, where none of the remaining dice are 2s or 6s. This leaves:

1, 3: solved by 6 * 6 * 3 - 6 / 3 + 1
3, 5: solved by 6 * 6 * 3 - 6 / 3 - 5

So if there are three 6s, the other three dice must be 1s, 4s, or 5s or there is a solution.

With three 6s, a 4, and a 5, we have (4 + 5) * (6 + 6), so we need a 1, 5, or 7 with one die and a 6. Therefore, there might not be a solution only if the die is a 4, so we have 4, 4, 5, 6, 6, 6. This is solved by (4 + 6) * (4 + 6) + 5 - 6.

Therefore, with three 6s, there cannot be a 4 and 5 at the same time, and we need to check cases with 1s and 4s, or with 1s and 5s (including none).

With three 6s, a 1, and a 4, we have 6 * (4 * 6 - 6), so we need a 1, 5, or 7 with one die and a 1. Since there is a 4, the die can be 1 or 4, and there is always a solution.

Therefore, with three 6s, there cannot be a 1 and a 4 at the same time, so we need to check the cases with three 1s, three 4s, and mixtures of 1s and 5s (including three 5s).

With three 6s, a 1, and a 5, we have (5 + 1) * (6 + 6 + 6), so we need a 1, 5, or 7 with the remaining die, which must be a 1 or a 5 - therefore, there is always a solution.

So we only need to check cases with three 6s and three 1s, 4s, or 5s.

1, 1, 1, 6, 6, 6: solved by 6 * 6 * 6 / (1 + 1) + 1
4, 4, 4, 6, 6, 6: solved by 6 * (4 * 6 - 6) - 4 / 4
5, 5, 5, 6, 6, 6: solved by 5 * 5 * 5 - 6 - 6 - 6

So there is always a solution with three or more 6s.


On to two 6s!

Two 6s can be used to get 36, so we just need a 3 and a 1, 5, or 7 to get a simple solution. Assuming we want to use two dice for each set, we have the following options:

(1, 2), (1, 3), (1, 4), (2, 5) for a 3
Any pair but (1, 3), (2, 4) and (3, 5) for a 1, 5, or 7.

So if there is a 1, a 3 is always possible unless the other three dice are 5s, which is solved with (5 + 5) * (5 + 6) - 6 - 1.

Therefore, if there is a 1, a 3 or a solution is possible, meaning that the other two dice are one of the three pairs that do not allow a 1, 5, or 7. This means that we have three possible sets of three dice, and a 'wild' one.

1, 1, 3 and wild: 1, 3 can be used to get 3, so the wild die must be a 3 as well (or a 1, 5, or 7 is possible). 3 * 6 * 6 - 3 + 1 + 1 is a solution.
1, 2, 4 and wild: 1, 2 can be used to get 3, so the wild die must be a 2. 1, 4 can be used to get 3, so the wild die must be a 4 as well. Therefore, there is a solution.
1, 3, 5 and wild: 1, 3 can be used to get 3, so the wild die must be a 3 (or a 1, 5, or 7 is possible). 3 * 6 * 6 - 5 * 1 is a solution.

So there is always a solution with two 6s if one of the other dice is a 1.

If there is a 2 and a 5, a 3 is possible, so the other two dice cannot give 1, 5, or 7. This means the following are options:

2, 2, 4, 5, 6, 6: solved by (2 + 4) * (2 * 6 + 6) - 5
2, 3, 5, 5, 6, 6: solved by 3 * 6 * 6 - 2 + 5 / 5

So there cannot be a 2 and 5 if there are two 6s.

If there is a 3, then we have 3 * 6 * 6 = 108, so we need the remaining three dice to give 1, 5, or 7. None of these dice can be 1s, and there cannot be 2s and 5s at the same time, so the following options exist:

2, 2, 2: solved by 2 - 2 / 2
2, 2, 3: solved by 2 + 2 + 3
2, 2, 4: solved by 4 + 2 / 2
2, 3, 3: solved by 2 - 3 / 3
2, 3, 4: solved by 2 + 3 - 4
2, 4, 4: solved by 2 - 4 / 4
3, 3, 3: no solution for 1, 5, 7. Solved by 3 * 3 * (6 + 6) - 3 / 3
3, 3, 4: solved by 4 + 3 / 3
3, 3, 5: solved by 5 * 3 / 3
3, 4, 5: solved by 5 * (4 - 3)
3, 5, 5: solved by 5 + 5 - 3
4, 4, 4: solved by 4 + 4 / 4
4, 4, 5: solved by 5 * 4 / 4
4, 5, 5: solved by 4 + 5 / 5
5, 5, 5: solved by 5 + 5 - 5

Therefore, if there is a 3, there is always a solution.

The remaining combinations involve 2s, 4s, and 5s, noting that there cannot be 2s and 5s at the same time.

If there are two 4s, then we have (6 + 4) * (6 + 4), so the remaining two dice must not allow 1, 3, or 7. This means that the dice cannot be a pair, and since 5 - 4 = 1, this leaves 2, 4, 4, 4, 6, 6, which is solved by 4 * 4 * 6 + (4 + 6) / 2. Therefore, there is at most a single 4, which will be combined with either 2s or 5s.

The possible combinations are then:

2, 2, 2, 2, 6, 6: no solution
2, 2, 2, 4, 6, 6: solved by (6 * 6 * (4 + 2) - 2) / 2
4, 5, 5, 5, 6, 6: solved by 5 * 5 * 5 - 4 * 6 + 6
5, 5, 5, 5, 6, 6: solved by 5 * 5 * 5 - 5 * 6 + 6

So with two 6s, there are no solutions only when the other four dice are all 2s, for a total of 15 combinations.


And we're down to a single 6 and the other five dice can be anything from 1 to 5.

If there is a 3, then we can try to split the other four dice into two pairs, with values of 6 and 1, 5, or 7.

We can get 6 with (1, 5), (2, 3), (2, 4), and (3, 3).

We can get 1, 5 or 7 with any pair but (1, 3), (2, 4) and (3, 5).

1, 1, 3, 3, 5, 6 is solved by 3 * 3 * 6 * (1 + 1) - 5
1, 2, 3, 4, 5, 6 is solved by using 2, 4 to get 6 and 1, 5 to get 5
1, 3, 3, 5, 5, 6 is solved by 5 * 5 * (3 + 1) + 6 - 3

So there is always a solution if there is a 1 and a 5 when there is a 3 and a 6.

1, 2, 3, 3, 3, 6 is solved by using 3, 3 to get 6, and 1, 2 to get 1
2, 2, 3, 3, 4, 6 is solved by using 2, 4 to get 6, and 2, 3 to get 1
2, 3, 3, 3, 5, 6 is solved by using 3, 3 to get 6, and 2, 5 to get 7

So there is always a solution if there is a 2 and an extra 3 when there is a 3 and a 6 (two 3s total).

With 2, 4, since there are 3s in most cases, we only need to check 2, 2, 3, 4, 4, 6, which is solved by 4 * 4 * 6 + 2 + 2 + 3

With 3, 3, since there is a solution when there is a 2 and a 3, we only need to check the following:

1, 3, 3, 3, 3, 6: solved by (6 + 3 + 3) * 3 * 3 - 1
3, 3, 3, 3, 5, 6: solved by (6 + 3 + 3) * 3 * 3 - 5

So there is always a solution if there is a 3 and a 6, and two other dice can be combined to get 6.

If there is a 3, a 6, and a 2, then other dice must be 1s, 2s, or 5s. Since there cannot be a 1 and a 5, this leaves the following:

1, 1, 1, 2, 3, 6: no solution (120 cases)
1, 1, 2, 2, 3, 6: solved by 6 * 3 * 2 * (2 + 1) - 1
1, 2, 2, 2, 3, 6: solved by 6 * 3 * (2 + 2 + 2) - 1
2, 2, 2, 2, 3, 6: no solution (30 cases)
2, 2, 2, 3, 5, 6: solved by 6 * 3 * (2 + 2 + 2) - 5
2, 2, 3, 5, 5, 6: solved by 5 * 5 * 2 * 2 + 6 - 3
2, 3, 5, 5, 5, 6: solved by 3 * 5 * 5 + 5 * 6 + 2
3, 5, 5, 5, 5, 6: solved by (3 + 5 + 5) * 6 + 5 * 5

So there are 150 combinations that are not solutions if there is a 2, 3, 6.

If we have a 1, 3, 6, then the other dice cannot be 5s (and 2s have been tested as well). If the remaining three dice can be used to get a 6, then there is a solution, which always happens if there are two or more 3s.

1, 1, 1, 1, 3, 6: no solution (30 cases)
1, 1, 1, 3, 3, 6: solved by (1 + 1) * 3 * 3 * 6 - 1
1, 1, 1, 3, 4, 6: solved by (4 + 1 + 1) * 3 * 6 - 1
1, 1, 3, 3, 4, 6: solved by 3 * (4 + 3 - 1) * 6 - 1
1, 1, 3, 4, 4, 6: solved by 4 * 4 * 6 + 3 + 1 + 1
1, 3, 3, 4, 4, 6: solved by 4 * 4 * 6 + 3 + 3 + 1

Which leaves 30 combinations that are not solutions if there is a 1, 3, 6.

With 3, 3, 6, the other dice must be 4s or 5s. This gives the following:

3, 3, 4, 4, 4, 6: solved by 4 * 4 * 6 + 4 + 3 / 3
3, 3, 4, 4, 5, 6: solved by (6 + 5) * 3 * 3 + 4 + 4
3, 3, 4, 5, 5, 6: solved by 4 * 5 * 5 + (3 + 3) / 6
3, 3, 5, 5, 5, 6: solved by (6 + 5) * (5 + 5) - 3 * 3

So that leaves the following combinations to check:

3, 4, 4, 4, 4, 6: solved by 4 * 4 * 6 + 4 + 4 - 3
3, 4, 4, 4, 5, 6: solved by 4 * 5 * 6 - 4 * 4 - 3
3, 4, 4, 5, 5, 6: solved by 4 * 5 * 5 + 3 + 4 - 6
3, 4, 5, 5, 5, 6: solved by 4 * 5 * 5 + 3 * (6 - 5)
3, 5, 5, 5, 5, 6: solved by 5 * 5 * (5 + 5 - 6) + 3

In all, that's 180 combinations that are not solutions if there is a 3 and a 6.

We can now study cases with one 6 and no 3s. (Current running total: 3827 combinations that are not successes, 6144 combinations left to check)

If there are two or more 4s, then 4 * 4 * 6 means that there is a solution if the remaining three dice give a 5, 7, or 11.

1, 1, 1: solved by (6 - 1) * (4 + 1) * 4 + 1
1, 1, 2: solved by (6 * 2 + 1) * (4 + 4) - 1
1, 1, 4: 4 + 1 * 1 = 5
1, 1, 5: 5 * 1 * 1 = 5
1, 2, 2: 1 + 2 + 2 = 5
1, 2, 4: 1 + 2 + 4 = 7
1, 2, 5: 1 * 2 + 5 = 7
1, 4, 4: 4 + 4 - 1 = 7
1, 4, 5: solved by (5 * 6 - 4) * 4 - 4 + 1
1, 5, 5: 5 + 5 + 1 = 11
2, 2, 2: no solution (60 cases)
2, 2, 4: 4 + 2 / 2 = 5
2, 2, 5: 5 * 2 / 2 = 5
2, 4, 4: no solution (30 cases)
2, 4, 5: 2 + 4 + 5 = 11
2, 5, 5: solved by 5 * 5 * 4 + (2 + 4) / 6
4, 4, 4: 4 + 4 / 4 = 5
4, 4, 5: 5 + 4 - 4 = 5
4, 5, 5: 4 + 5 / 5 = 5
5, 5, 5: 5 + 5 - 5 = 5

So there are 90 cases in all with two or more 4s where there is no solution (2, 2, 2, 4, 4, 6 and 2, 4, 4, 4, 4, 6)

With a 6 and one 4, the remaining four dice are 1s, 2s, and 5s.

1, 1, 1, 1, 4, 6: no solution (30 cases)
1, 1, 1, 2, 4, 6: no solution (120 cases)
1, 1, 1, 4, 5, 6: solved by 6 * (5 + 1) * (4 - 1) - 1
1, 1, 2, 2, 4, 6: solved by 6 * (4 + 2) * (2 + 1) - 1
1, 1, 2, 4, 5, 6: solved by 6 * 4 * (2 + 1 + 1) + 5
1, 1, 4, 5, 5, 6: solved by 4 * 5 * 5 + 6 + 1 * 1
1, 2, 2, 2, 4, 6: solved by (6 * 4 + 2) * 2 * 2 - 1
1, 2, 2, 4, 5, 6: solved by (5 * 6 - 4) * 2 * 2 - 1
1, 2, 4, 5, 5, 6: solved by 4 * 5 * 5 + 6 + 2 - 1
1, 4, 5, 5, 5, 6: solved by 4 * 5 * 5 + (1 + 5) / 6
2, 2, 2, 2, 4, 6: no solution (30 cases)
2, 2, 2, 4, 5, 6: solved by 6 * 4 * 2 * 2 + 5 + 2
2, 2, 4, 5, 5, 6: solved by 4 * 5 * 5 + 6 + 2 / 2
2, 4, 5, 5, 5, 6: solved by 4 * 5 * 5 + 2 + 5 - 6
4, 5, 5, 5, 5, 6: solved by 4 * 5 * 5 + 6 + 5 / 5

So if there are any 4s, there are 270 cases without a solution (90 with two or more 4s, 180 with one 4) - meaning 4097 combinations are not successes so far.


We're down to cases with one 6, and the other dice are 1s, 2s, or 5s.

1, 1, 1, 1, 1, 6: no solution (6 cases)
1, 1, 1, 1, 2, 6: no solution (30 cases)
1, 1, 1, 1, 5, 6: no solution (30 cases)
1, 1, 1, 2, 2, 6: no solution (60 cases)
1, 1, 1, 2, 5, 6: solved by 6 * (5 + 1) * (2 + 1) - 1
1, 1, 1, 5, 5, 6: solved by (6 - 1) * 5 * (5 - 1) + 1
1, 1, 2, 2, 2, 6: no solution (60 cases)
1, 1, 2, 2, 5, 6: solved by (6 - 1) * 5 * 2 * 2 + 1
1, 1, 2, 5, 5, 6: solved by 5 * 5 * (6 - 2) + 1 * 1
1, 1, 5, 5, 5, 6: solved by 5 * 5 * (5 - 1) + 6 + 1
1, 2, 2, 2, 2, 6: no solution (30 cases)
1, 2, 2, 2, 5, 6: solved by (5 + 2 + 2) * 6 * 2 - 1
1, 2, 2, 5, 5, 6: solved by 5 * 5 * (6 - 2) + 2 + 1
1, 2, 5, 5, 5, 6: solved by 5 * 5 * (5 - 1) + 6 / 2
1, 5, 5, 5, 5, 6: solved by 5 * 5 * (5 - 1) + 6 - 5
2, 2, 2, 2, 2, 6: no solution (6 cases)
2, 2, 2, 2, 5, 6: solved by 6 * 2 * 2 * 2 * 2 + 5
2, 2, 2, 5, 5, 6: solved by 5 * 5 * 2 * 2 + 6 / 2
2, 2, 5, 5, 5, 6: solved by 5 * 5 * 2 * 2 + 6 - 5
2, 5, 5, 5, 5, 6: solved by 5 * 5 * (6 - 2) + 5 / 5
5, 5, 5, 5, 5, 6: solved by (5 + 5) * (5 + 5) + 6 - 5

Which gives 222 cases which are not solutions.

The grand total (barring new solutions - I've found a few on review) is therefore 4319 / 46656 cases are not solutions.


Final results, assuming I haven't missed any solutions: with 6 dice, a level 9 spell can be successfully cast 90% of the time.

So you need a few more dice to hit 99+%, but the odds of success appear to go up as spell levels go down, so 6 ranks are likely to be enough for a hefty part of your career. The jump from single digit percentages to 90% is rather precipitous.

[Chronos]

The person who took the feat signed up for it, but the rest of the players didn't. It's one thing to have to wait a little longer at the table because another player is figuring out how much he should Power Attack for: That's the sort of thing that you expect to maybe have to wait for. It's another to have to wait longer because you have to wait for another player to play some silly game about arranging plus signs and parentheses.

[Snowbluff]

Sometimes when I see these horribly overpowered combinations, or confusingly-worded abilities, I half suspect that the authors who created them were just trying to foist them off on their own DMs for purposes of their own personal powergaming. See, it's published! It's a real rule! I totally get to use it in game!This is how I feel about homebrew a vast majority of the time. :smalltongue:

I don't think the dev is smart enough for that though.

[3WhiteFox3]

The person who took the feat signed up for it, but the rest of the players didn't. It's one thing to have to wait a little longer at the table because another player is figuring out how much he should Power Attack for: That's the sort of thing that you expect to maybe have to wait for. It's another to have to wait longer because you have to wait for another player to play some silly game about arranging plus signs and parentheses.

See, I've been using it in a small home game and we haven't had any problems with speed. I actually tend to take less time with my turns than some of the other players. Especially if the GM allows me to roll the dice during other people's turns and do the calculations there. I'm not even that great at math either (it's actually one of my weaker subjects). 1 or 2 minutes on average, and that's not nearly as much time as I've seen it take for some of the slower players to calculate damage or what spell they're going to use, etc...

I especially like it, because it gives me something to do when the other players are taking forever with their turns, I roll the dice and make the quick calculations, and voila it works. Now, I do feel that in most groups the feats are going to be problematic, but that's because of power-level, not math.

[Chronos]

But that's time that you're spending in addition to whatever time you'd be spending on your spell effects, whatever they are.

[3WhiteFox3]

But that's time that you're spending in addition to whatever time you'd be spending on your spell effects, whatever they are.

Yeah, but spell effects are much easier than trying to keep up with (for example) multiple attacks, temporary bonuses and penalties, ability score modifications, damage rolls, damage mods, etc... Casters almost always take less time than martials because they have much less to calculate and most of their calculations can be done ahead of time. It's basically just Spell DCs and Spell Slots, that's it. And while some spells take longer to apply, most are in fact very easy. Besides, if a DM has a problem with it, just let them roll when it's not their turn. Problem solved.

[grarrrg]

Ran some numbers to see how Calculating Mind (d8's) stack up.

These were fairly easy, as I only need to worry about possibilities that have a 7 or 8, as we've already done all the work on 1-through-6.
(I was more fast than careful, so my numbers may be off a touch (due to the method used, the actual success rate may be higher, but won't be lower > assume fail, prove pass).

3 Dice, Level 1 spells:
Only d6's: 91.67% to pass (198/216)
Add d8's: 83.59% to pass (428/512)

3 Dice, Level 2 spells:
d6's: 52.78% pass (114/216)
d8's: 56.25% pass (288/512)

3 Dice, Level 3 spells:
d6's: 16.67% pass (36/216)
d8's: 27.54% pass (141/512)

3 Dice, Level 5 spells:
d6's: 0%
d8's: 10.55 (54/512)

4 Dice, Level 1 spells:
d6's: 100%
d8's: 100%


Findings: Using d8's at the high end (more dice) actually hurts chances (would need to run d8's at more "100%" levels to see if any failures appear).
Also worth noting that it is impossible for d8's to reduce the dice needed for 100% pass, as it doesn't eliminate any current results that fail at lower dice (i.e. all 1's).

Using d8's appears to help if using few dice, and occasionally enables a chance that wasn't there previously (5th level using 3 dice).
BUT, if someone is using this feat chances are they are pumping Know-(Engineering), thus negating any potential usefulness.

Conclusion: Calculating Mind is garbage. The only functional use would be if you somehow can only afford, like, 5 ranks max in Know(Engineering). And in that case you are probably better off just taking 2 entirely different feats.

[meschlum]

And, even more to the point, you could get four metamagic feats for which Sacred Geometry works by taking it twice, rather than only two metamagics and possibly increased odds.

Running six dice for level 9 is giving a fairly high chance of success so far (the calculations are not complete), so the possibly improved odds you get from Calculating Mind are likely to be even less relevant. This assumes that strange things do not happen to the odds on the way (more dice = higher chance of success and lower level = higher chance of success).

Having three more dice (d6s) is always better than the original, since the three dice can cancel out via grrag's method, and there may be more combinations available with them which increase the odds. Having one or two more dice may not be better, though it seems very likely to be.

Lower levels aim for lower target numbers, so fewer dice are involved in getting close. The frequency and ability to reach said target numbers is variable, so it's not absolutely guaranteed to give better odds. But, again, it seems likely.

[meschlum]

The urge to math struck me, and I finished the 6 die tests.

Assuming I haven't missed anything, level 9 spells can be cast ~90% of the time with 6 ranks. So it's not quite 99% for level 9, but since the chance of success goes up as spell levels go down, 6 ranks look like they'll be plenty for a solid portion of your casting career (even more so if you're a bard or some such).

Since adding 3 dice lets you use all the previous solutions, going to 9 dice means you're still at 90%, except you now have tree extra dice with which to try and solve the (rare) cases that didn't work initially. So I'd suggest 9 dice seem a reasonable quantity for 99% success, requiring a bit of work from time to time.